D. Substring

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a graph with n nodes and m directed edges. One lowercase letter is assigned to each node. We define a path’s value as the number of the most frequently occurring letter. For example, if letters on a path are “abaca”, then the value of that path is 3. Your task is find a path whose value is the largest.

Input

The first line contains two positive integers n, m (1 ≤ n, m ≤ 300 000), denoting that the graph has n nodes and m directed edges.

The second line contains a string s with only lowercase English letters. The i-th character is the letter assigned to the i-th node.

Then m lines follow. Each line contains two integers x, y (1 ≤ x, y ≤ n), describing a directed edge from x to y. Note that x can be equal to y and there can be multiple edges between x and y. Also the graph can be not connected.

Output

Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.

Examples

input

5 4
abaca
1 2
1 3
3 4
4 5

output

3

input

6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4

output

-1

input

10 14
xzyzyzyzqx
1 2
2 4
3 5
4 5
2 6
6 8
6 5
2 10
3 9
10 9
4 6
1 10
2 8
3 7

output

4

Note

In the first sample, the path with largest value is 1 → 3 → 4 → 5. The value is 3 because the letter ‘a’ appears 3 times.

题目链接：http://codeforces.com/problemset/problem/919/D

题意:给定一个字符串，第i个字符代表第i个节点，求最长的路径（当前路径的最大权值为该路径中出现次数最多的字符），若结果无限大则输出-1

思路：拓扑排序+状态转移，不能遍历所有的节点则输出-1，否则每遍历一个节点则更新以当前节点为路径终点中各个字符出现的最大值。

代码如下：

#include <bits/stdc++.h>
using namespace std;

const int N = 3e5+10;
int n,m,f[N][30],head[N],cnt,deg[N];
char a[N];
struct Edge{
    int to,next;
}edge[N*2];

void add(int u, int v){
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    head[u] = cnt ++;
}

void toposort(){
    queue<int>q;
    for(int i = 1; i <= n; i ++){
        if(!deg[i]){
            q.push(i);
            f[i][a[i-1]-'a'] ++;
        }
    }
    int cnt = 0;
    while(!q.empty()){
        int u = q.front(); q.pop();
        cnt ++;
        for(int i = head[u]; ~i; i = edge[i].next){
            int v = edge[i].to;
            deg[v] --;
            if(!deg[v]) q.push(v);
            for(int j = 0; j < 26; j ++) f[v][j] = max(f[v][j],f[u][j] + (a[v-1]-'a' == j));
        }
    }
    if(cnt < n) printf("-1\n");
    else{
        int ans = 0;
        for(int i = 1; i <= n; i ++){
            for(int j = 0; j < 26; j ++){
                ans = max(ans,f[i][j]);
            }
        }
        printf("%d\n",ans);
    }
}

int main(){
    scanf("%d%d%s",&n,&m,a);
    memset(head,-1,sizeof(head));
    cnt = 0;
    int u,v;
    for(int i = 0; i < m; i ++){
        scanf("%d%d",&u,&v);
        deg[v] ++;
        add(u,v);
    }
    toposort();
    return 0;
}