Masha and two friends

                                                                            time limit per test:1 second

                                                                            memory limit per test:256 megabytes

                                                                            input standard input

                                                                            output standard output

Recently, Masha was presented with a chessboard with a height of n and a width of m

The rows on the chessboard are numbered from 1 to n from bottom to top. The columns are numbered from 1 to m from left to right. Therefore, each cell can be specified with the coordinates (x,y), where x is the column number, and y is the row number (do not mix up).

Let us call a rectangle with coordinates (a,b,c,d) a rectangle lower left point of which has coordinates (a,b) , and the upper right one — (c,d) .

The chessboard is painted black and white as follows:

An example of a chessboard

Masha was very happy with the gift and, therefore, invited her friends Maxim and Denis to show off. The guys decided to make her a treat — they bought her a can of white and a can of black paint, so that if the old board deteriorates, it can be repainted. When they came to Masha, something unpleasant happened: first, Maxim went over the threshold and spilled white paint on the rectangle (x1,y1,x2,y2) . Then after him Denis spilled black paint on the rectangle (x3,y3,x4,y4) .

To spill paint of color colorcolor onto a certain rectangle means that all the cells that belong to the given rectangle become colorcolor . The cell dyeing is superimposed on each other (if at first some cell is spilled with white paint and then with black one, then its color will be black).

Masha was shocked! She drove away from the guests and decided to find out how spoiled the gift was. For this, she needs to know the number of cells of white and black color. Help her find these numbers!

Input

The first line contains a single integer tt (1≤t≤1031≤t≤103 ) — the number of test cases.

Each of them is described in the following format:

The first line contains two integers nn and mm (1≤n,m≤10^9 ) — the size of the board.

The second line contains four integers x1 , y1 , x2 , y2 (1≤x1≤x2≤m,1≤y1≤y2≤n ) — the coordinates of the rectangle, the white paint was spilled on.

The third line contains four integers x3 , y3 , x4 , y4 (1≤x3≤x4≤m,1≤y3≤y4≤n ) — the coordinates of the rectangle, the black paint was spilled on.

Output

Output tt lines, each of which contains two numbers — the number of white and black cells after spilling paint, respectively.

Input

5
2 2
1 1 2 2
1 1 2 2
3 4
2 2 3 2
3 1 4 3
1 5
1 1 5 1
3 1 5 1
4 4
1 1 4 2
1 3 4 4
3 4
1 2 4 2
2 1 3 3

Output

0 4
3 9
2 3
8 8
4 8

Note

Explanation for examples:

The first picture of each illustration shows how the field looked before the dyes were spilled. The second picture of each illustration shows how the field looked after Maxim spoiled white dye (the rectangle on which the dye was spilled is highlighted with red). The third picture in each illustration shows how the field looked after Denis spoiled black dye (the rectangle on which the dye was spilled is highlighted with red).

In the first test, the paint on the field changed as follows:

In the second test, the paint on the field changed as follows:

In the third test, the paint on the field changed as follows:

 In the fourth test, the paint on the field changed as follows:

In the fifth test, the paint on the field changed as follows:

題意:

給你一個n*m的棋盤，最初(1,1)上為白色，而且每個相鄰的塊顏色都不同。
之後有兩次操作，
第一次操作給出x1,y2,x2,y2
將(x1,y1,x2,y2)這個矩形塗為白色
第二次操作給出x3,y3,x4,y4
將(x3,y3,x4,y4)這個矩形塗為黑色
後塗得會覆蓋之前的顏色。問最終的棋盤上黑色和白色的個數

題解:

1.先呼叫函式求出所給區域中白色快的數量

2.未考慮重合部分白色塊的數量=總的白色塊數量+(第一次選定區域的整個部分-第一次選定區域中的白色部分)(因為全部被染成白色了)-第二次選定區域中的白色部分(因為被染成黑色了)

3.考慮重合部分

一共有五種可能，其中能重合的只有四種，畫畫圖就會發現關係

 xx1=max(x1,x3);
 xx2=min(x2,x4);
 yy1=max(y1,y3);
 yy2=min(y2,y4);

要想保證重合需要滿足(xx1<=xx2 && yy1<=yy2) 

答案=未考慮重合部分白色塊的數量-整個重疊部分+重疊部分中白色的部分

注意:這個題有個坑點，對於初始狀態白色塊與黑色快的數量不相同 如下圖所示

小技巧： ans=((x2-x1+1)*(y2-y1+1)+1)/2; 向上取整的操作 cile(m/n)=(m+n-1)/n;

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <utility>
#include <set>
#include <bitset>
#include <vector>
#define pi acos(-1.0)
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fLL
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
int min3(int a,int b,int c){return min(min(a,b),c);}
int max3(int a,int b,int c){return max(max(a,b),c);}
int gcd(int x, int y){if(y==0)return x;return gcd(y, x%y);}
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;};

ll ws(ll x1,ll y1,ll x2,ll y2)///返回白色的數量
{
    ll ans=0;
    if((x2-x1+1)%2==0 || (y2-y1+1)%2==0)
        ans=(x2-x1+1)*(y2-y1+1)/2;
    else if((x1+y1)%2)/// 表示左下角(x1,y1)為黑色的
        ans=(x2-x1+1)*(y2-y1+1)/2;
    else ans=(x2-x1+1)*(y2-y1+1) - (x2-x1+1)*(y2-y1+1)/2;  ///白的=總的-黑的;
    /// ans=((x2-x1+1)*(y2-y1+1)+1)/2; 向上取整的操作 cile(m/n)=(m+n-1)/n;
    return ans;

}

int main()
{
    int t;
    ll n,m;
    ll x1,y1,x2,y2;
    ll x3,y3,x4,y4;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld",&n,&m);
        scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2);
        scanf("%lld%lld%lld%lld",&x3,&y3,&x4,&y4);

        ///應用容斥原理
        ll ans1;
        ans1=ws(1,1,n,m)+((x2-x1+1)*(y2-y1+1)-ws(x1,y1,x2,y2))-ws(x3,y3,x4,y4);


        ///白色=總的白色+(第一次選定區域的整個部分-第一次選定區域中的白色部分)(因為全部被染成白色了)-第二次選定區域中的白色部分(因為被染成黑色了)這是未考慮重合部分

        ///考慮重合部分
        ll xx1=max(x1,x3);
        ll xx2=min(x2,x4);
        ll yy1=max(y1,y3);
        ll yy2=min(y2,y4);


        if(xx1<=xx2 && yy1<=yy2)
        {
            ans1-=(xx2-xx1+1)*(yy2-yy1+1);///減去整個重疊部分+重疊部分中白色的部分
            ans1+=ws(xx1,yy1,xx2,yy2);
        }
        ll ans2=n*m-ans1;

        printf("%lld %lld\n",ans1,ans2);

    }
	return 0;

}