C. Vasya and String time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.

Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve?

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.

The second line contains the string, consisting of letters 'a' and 'b' only.

Output

Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k

characters.

Examples Input

4 2
abba

Output

4

Input

8 1
aabaabaa

Output

5

Note

In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".

In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".

題意：給你一個字串，只有a，b，最多k此操作，每次操作可以任意改變一個字元，求最長的連續相同字元長度

思路：直接用佇列計入要改的字元的位置， 然後跑一邊就可以了，複雜度O(2 * n);

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int inf = 2147483647;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const int mod = 1000000007;
typedef long long LL;
#pragma comment(linker, "/STACK:102400000,102400000")
//freopen("in.txt","r",stdin); //輸入重定向,輸入資料將從in.txt檔案中讀取
//freopen("out.txt","w",stdout); //輸出重定向,輸出資料將儲存在out.txt檔案中cin
char s[100005];
queue<int>sq;
int main()
{
	int n, k, i, j;
	while (cin >> n >> k)
	{
		cin >> s + 1;
		int mmax = 0;
		int last = 1;
		int ans = 0;
		while (!sq.empty())
			sq.pop();
		for (i = 1; i <= n; ++i)
		{
			if (s[i] == 'b')
			{			
				if (sq.size() < k)
				{
					sq.push(i);
					mmax = max(mmax, i - last + 1);
				}
				else 
				{
					if (sq.size() != 0)
					{
						last = sq.front() + 1;
						sq.pop();
						sq.push(i);
					}
					else
						last = i + 1;
				}				
				ans = 1;
			}
			else
				mmax = max(mmax, i - last + 1);
		}
		//cout << ans << endl;
		if (!ans)
			mmax = n;
		while (!sq.empty())
			sq.pop();
		ans = 0;
		last = 1;
		for (i = 1; i <= n; ++i)
		{
			if (s[i] == 'a')
			{
				//mmax = max(mmax, i - 1 - last + 1);
				if (sq.size() < k)
				{
					sq.push(i);
					mmax = max(mmax, i - last + 1);
				}
				else
				{
					if (sq.size() != 0)
					{
						last = sq.front() + 1;
						sq.pop();
						sq.push(i);
					}
					else
						last = i + 1;
				}
				ans = 1;
			}
			else
				mmax = max(mmax, i - last + 1);
		}
		if (!ans)
			mmax = n;
		cout << mmax << endl;

	}
}