Codeforces Round #460 (Div. 2) D Substring

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.

有n个节点和m条边，图是有向的，每个节点有一个小写字符，求一条路径中出现相同字符最大的数量。

dfs+dp问题，dp[i][j]表示以i节点开始的路径中字母'a'+j出现的最大次数。然后j遍历0-26，就可以求出最大值了。

#include <bits/stdc++.h>
using namespace std;
const int N = 3e5+10;
char str[N];
int n, m, x, y, ans;
vector<int> vs[N];
int vis[N];
int dp[N][26];

void dfs(int u) {
	vis[u] = 1;
	dp[u][str[u]-'a'] = 1;
	for(int i = 0; i < vs[u].size(); i ++) {
		int v = vs[u][i];
		if(vis[v] == 1) {
			printf("-1
");
			exit(0);
		} else{
			if(!vis[v]) dfs(v);
			for(int j = 0; j < 26; j ++) {
				dp[u][j] = max(dp[u][j],dp[v][j]+(str[u]-'a' == j));
			}
		}
	}
	vis[u] = 2;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> m;
	cin >> str+1;
	for(int i = 0; i < m; i ++) {
		cin >> x >> y;
		vs[x].push_back(y);
	}
	memset(vis,0,sizeof(vis));
	memset(dp,0,sizeof(dp));
	for(int i = 1; i <= n; i ++) {
		if(!vis[i]) {
			dfs(i);
			for(int j = 0; j < 26; j ++) {
				ans = max(ans,dp[i][j]);
			}
		}
	}
	cout << ans << endl;
	return 0;
}