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.
from collections import defaultdict, deque, Counter
import sys
import bisect
import math

input = sys.stdin.readline
mod = 1000000007

n ,m = input().split()
n = int(n)
m = int(m)
l = list(input())
degree = [0 for i in range(n)]
graph = defaultdict(list)
for i in range(m):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    graph[a].append(b)
    degree[b] += 1
q = deque()
for i in range(n):
    if degree[i] == 0:
        q.append(i)
count = 0
ans = 0
# print(degree)
dp = [[0 for i in range(26)] for i in range(n)]
while count < n and q:
    x = q.popleft()
    count += 1
    dp[x][ord(l[x]) - 97] += 1
    for i in graph[x]:
        for j in range(26):
            dp[i][j] = max(dp[i][j], dp[x][j])
        degree[i] -= 1
        if degree[i] == 0:
            q.append(i)
if count != n:
    print(-1)
else:
    ans = 0
    for i in range(n):
        ans = max(ans, max(dp[i]))
    print(ans)