919_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.
```python
from collections import defaultdict
import sys
from sys import stdin
def check(a):
    for i in a:
        if i:
            return True
    return False
def change(x):
    return ord(x)-ord('a')  
def solve():
    n, m = map(int, stdin.readline().split())
    s = input()
    d = {}
    for i in range(n):
        d[i+1] = s[i]
    g = defaultdict(list)#[[] for _ in range(n+1)]
    ind = [0 for _ in range(n+1)]
    for _ in range(m):
        x, y = map(int, stdin.readline().split())
        g[x].append(y)
        ind[y]+=1
    q = []
    dp = [[0 for _ in range(27)]for _ in range(n + 1)]
    for i in range(1, n+1):
        if ind[i]==0:
            q.append(i)
            dp[i][change(d[i])] = 1
    for i in q:
        for j in g[i]:
            for k in range(26):
                if change(d[j])==k:
                    dp[j][k] = max(dp[i][k]+1, dp[j][k])
                else:
                    dp[j][k] = max(dp[i][k], dp[j][k])
            ind[j]-=1
            if ind[j]==0: q.append(j)
    if check(ind):
        print(-1)
    else:
        ans = 0
        for i in dp:
            for j in i:
                ans = max(ans, j)
        print(ans)
t = 1   
#t = inp()
for _ in range(t):
    solve()

```