Ksusha the Squirrel is standing at the beginning of a straight road, divided into n sectors. The sectors are numbered 1 to n, from left to right. Initially, Ksusha stands in sector 1.

Ksusha wants to walk to the end of the road, that is, get to sector n. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn’t want to stand in sectors that have rocks.

Ksusha the squirrel keeps fit. She can jump from sector i to any of the sectors i + 1, i + 2, …, i + k.

Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)?

Input
The first line contains two integers n and k (2 ≤ n ≤ 3·105, 1 ≤ k ≤ 3·105). The next line contains n characters — the description of the road: the i-th character equals “.”, if the i-th sector contains no rocks. Otherwise, it equals “#”.

It is guaranteed that the first and the last characters equal “.”.

Output
Print “YES” (without the quotes) if Ksusha can reach the end of the road, otherwise print “NO” (without the quotes).

Examples
Input
2 1
…
Output
YES
Input
5 2
.#.#.
Output
YES
Input
7 3
.#.###.
Output
NO
又挺長時間沒有寫部落格了，今天晚上補一波。
題目很簡單，就是走格子，但是不能走到‘#’，每次可以走到i+1,i+2…i+k，要知道，如果i+1不行的話，那麼i+2,3也不行，所以找到最小的哪一步就好了。
程式碼如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int maxx=3e5+10;
char s[maxx];
int n,k;

int main()
{
	while(cin>>n>>k)
	{
		cin>>s;
		int i,j;
		int flag=0;
		for(i=0;i!=n-1;)
		{
			for(j=1;j<=k;j++)
			{
				if(s[i+j]=='.')
				{
					i+=j;
					break; 
				}
			}
			if(j>k) 
			{
				flag=1;
				break;
			}
		}
		if(flag) cout<<"NO"<<endl;
		else cout<<"YES"<<endl;
	}
}

努力加油a啊，(^o)/~