Given a string S and two integers L and M, we consider a substring of S as “recoverable” if and only if
  (i) It is of length M*L;
  (ii) It can be constructed by concatenating M “diversified” substrings of S, where each of these substrings has length L; two strings are considered as “diversified” if they don’t have the same character for every position.

Two substrings of S are considered as “different” if they are cut from different part of S. For example, string "aa" has 3 different substrings "aa", "a" and "a".

Your task is to calculate the number of different “recoverable” substrings of S.

 

 

The input contains multiple test cases, proceeding to the End of File.

The first line of each test case has two space-separated integers M and L.

The second ine of each test case has a string S, which consists of only lowercase letters.

The length of S is not larger than 10^5, and 1 ≤ M * L ≤ the length of S.

 

For each test case, output the answer in a single line.

 

 

 

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<algorithm>

#include<queue>

#include<cmath>

#include<map>

using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")

#define ls i<<1

#define rs ls | 1

#define mid ((ll+rr)>>1)

#define pii pair<int,int>

#define MP make_pair

typedef long long LL;

typedef unsigned long long ULL;

const long long INF = 1e18+1LL;

const double pi = acos(-1.0);

const int N = 5e5+, MM = 1e3+,inf = 2e9;

const LL mod = 10000019ULL;

inline LL read()

{

    LL x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

map<ULL,int  > s;

ULL bhas[N],has[N],sqr[N];

int M,L;

char sa[N];

int vis[N];

int main() {

    sqr[] = ;

    for(int i = ; i < N; ++i) sqr[i] = sqr[i-] * mod;

    while(scanf("%d%d",&M,&L)!=EOF) {

        scanf("%s",sa+);

        int n = strlen(sa+);

        has[] = ;

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

            has[i] = has[i-] * mod + sa[i] - 'a' + ;

        }

        int ans = ;

        for(int i = ; i <= L && i + M * L -  <= n; ++i) {

            int cnt = ;

            s.clear();

            int ll = ,rr = ;

            for(int j = i; j + L -  <= n; j += L) {

                int l = j, r = j + L - ;

                ULL now = has[r] - has[l-]*sqr[L];

                if(s[now] == ){

                    bhas[++rr] = now;

                    if(rr - ll +  >= M)ans+=;

                    s[now] = ;

                    continue;

                }

                else {

                    while(ll <= rr && bhas[ll]!=now) {

                        s[bhas[ll++]] = ;

                    }

                    s[bhas[ll++]] = ;

                    bhas[++rr] = now;

                    if(rr - ll +  >= M)ans+=;

                    s[now] = ;

                }

            }

        }

        printf("%d\n",ans);

    }

    return ;

}