Codeforces 1105B：Zuhair and Strings（字符串水题）

time limit per test: 1 second

memory limit per test: 256 megabytes

input: standard input

output: standard output

Given a string sss of length nnn and integer k (1≤k≤n)k\ (1≤k≤n)k (1≤k≤n). The string sss has a level xxx, if xxx is largest non-negative integer, such that it’s possible to find in sss:

xxx non-intersecting (non-overlapping) substrings of length kkk,
all characters of these x substrings are the same (i.e. each substring contains only one distinct character and this character is the same for all the substrings).

A substring is a sequence of consecutive (adjacent) characters, it is defined by two integers iii and j (1≤i≤j≤n)j\ (1≤i≤j≤n)j (1≤i≤j≤n), denoted as s[i…j]s[i…j]s[i…j] = “sisi+1…sjs_is_{i+1}…s_jsisi+1…sj”.

For example, if k=2k=2k=2, then:

the string “aabb” has level 111 (you can select substring “aa”),

the strings “zzzz” and “zzbzz” has level 222 (you can select two non-intersecting substrings “zz” in each of them),

the strings “abed” and “aca” have level 000 (you can’t find at least one substring of the length k=2k=2k=2 containing the only distinct character).

Zuhair gave you the integer kkk and the string s of length nnn. You need to find xxx, the level of the string sss.

Input

The first line contains two integers nnn and k (1≤k≤n≤2⋅105)k\ (1≤k≤n≤2⋅10^5)k (1≤k≤n≤2⋅105) — the length of the string and the value of kkk.

The second line contains the string sss of length n consisting only of lowercase Latin letters.

Output

Print a single integer xxx — the level of the string.

Examples

input

8 2

aaacaabb

output

2

input

2 1

ab

output

1

input

4 2

abab

output

0

Note

In the first example, we can select 222 non-intersecting substrings consisting of letter ‘a’: “(aa)ac(aa)bb”, so the level is 222.

In the second example, we can select either substring “a” or “b” to get the answer 111.

题意

给出一个长度为nnn的字符串，在字符串中查找连续出现kkk次的字母，中间不能有重叠，连续出现kkk次的字母最多出现了几次

AC代码

暴力查找即可，注意k=1k=1k=1的情况

#include <stdio.h>

#include <string.h>

#include <iostream>

#include <algorithm>

#include <math.h>

#include <limits.h>

#include <map>

#include <stack>

#include <queue>

#include <vector>

#include <set>

#include <string>

#include <time.h>

#define ll long long

#define ull unsigned long long

#define ms(a,b) memset(a,b,sizeof(a))

#define pi acos(-1.0)

#define INF 0x7f7f7f7f

#define lson o<<1

#define rson o<<1|1

#define bug cout<<"-------------"<<endl

#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"

const double E=exp(1);

const int maxn=1e6+10;

const int mod=1e9+7;

using namespace std;

char ch[maxn];

int a[maxn];

bool cmp(int a,int b)

{

	return a>b;

}

int main(int argc, char const *argv[])

{

	ios::sync_with_stdio(false);

	int n,k;

	cin>>n>>k;

	cin>>ch;

	int res=1;

	for(int i=0;i<n;i++)

	{

		if(k==1)

			a[ch[i]-'a']++;

		else

		{

			if(ch[i]==ch[i+1])

				res++;

			else

				res=1;

			if(res==k)

			{

				a[ch[i]-'a']++;

				res=1;

				i+=1;

			}

		}

	}

	sort(a,a+26,cmp);

	cout<<a[0]<<endl;

	return 0;

}