Given n different objects, you want to take k of them. How many ways to can do it?

For example, say there are 4 items; you want to take 2 of them. So, you can do it 6 ways.

Take 1, 2

Take 1, 3

Take 1, 4

Take 2, 3

Take 2, 4

Take 3, 4

Input

Input starts with an integer T (≤ 2000), denoting the number of test cases.

Each test case contains two integers n (1 ≤ n ≤ 10⁶), k (0 ≤ k ≤ n).

Output

For each case, output the case number and the desired value. Since the result can be very large, you have to print the result modulo 1000003.

 費小馬定理求逆元

a/b=1mod( M );

只要 M 是一個素數，而且 b 不是 M 的倍數，就可以用一個逆元整數 b’，通過 a/b=a*b'*(mod M)來以乘換除

費馬小定理說，對於素數 M 任意不是 M 的倍數的 b，都有：b^(M-1)=1 (mod) M;

於是可以拆成：b*b^(M-2)=1(mod)M;

所以：a/b=a/b*(b*b^(M-2))=a*(b^(M-2))(mod M)

也就是說我們要求的逆元就是b^(M-2)(mod M)

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e6+5,mod=1000003;
typedef long long LL;
LL fact[maxn];//不明白為什麼這裡我用define定義maxn就不行 
void f()//打出階乘表 
{
    fact[1]=fact[0]=1;
    for(LL  i=2;i<maxn;i++)
    fact[i]=(i*fact[i-1])%mod;
}
LL niyuan(LL a,LL p)//快速冪
{
    LL res=1%mod;
	LL t=a%mod;
	while(p)
	{
		if(p%2)
		res=res*t%mod;
		t=t*t%mod;
		p=p/2;
	}
	return res;
}
LL c(int n,int k)
{
    LL fm=(fact[k]*fact[n-k])%mod;
    LL ans1=niyuan(fm,mod-2);//費馬小定理，求一個冪就好;
    return (ans1*fact[n])%mod;
}
int main()
{
    int t,n,k,kase=0;
    f();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        if(k*2>n)
        k=n-k;//組合數對稱性，減少計算；
        printf("Case %d: %lld\n",++kase,c(n,k));
    }
    return 0;
}