CodeForces 919D Substring

Description

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 \(n, m \left(1\le n,m \le 300000\right)\), 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 \(x, y \left(1 \le x,y \le n \right)\), 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 \(1 \rightarrow 3 \rightarrow 4 \rightarrow 5\). The value is \(3\) because the letter 'a' appears \(3\) times.

Solution

如果图中存在环，可以得到无限长的路径，所以答案为\(-1\)；否则，给定的图为DAG，记忆化搜索，求出每个字母在单条路径上的最大次数，即可得到答案。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 300011;
char s[maxn];
int deg[maxn], d[maxn];
vector<int> w[maxn];
int Find(int u, int c) {
  if (d[u] != -1) return d[u];
  d[u] = 0;
  for (int v : w[u]) {
    d[u] = max(d[u], Find(v, c));
  }
  d[u] += (s[u] == c + 'a');
  return d[u];
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    w[u].push_back(v);
    ++deg[v];
  }
  int ct = 0;
  queue<int> que;
  for (int i = 1; i <= n; ++i) {
    if (!deg[i])
      que.push(i), ++ct;
  }
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int v : w[u]) {
      if (--deg[v] == 0)
        que.push(v), ++ct;
    }
  }
  if (ct < n) {
    puts("-1");
  } else {
    int ans = 0;
    for (int c = 0; c < 26; ++c) {
      memset(d, 0xff, sizeof d);
      for (int i = 1; i <= n; ++i)
        ans = max(ans, Find(i, c));
    }
    printf("%d\n", ans);
  }
  return 0;
}