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.
import sys
from queue import Queue
from collections import defaultdict, deque

input = sys.stdin.readline

n, m = map(int, input().split())
s = list(input())

degree = [0 for _ in range(n)]
dp = [[0 for _ in range(26)] for _ in range(n)]

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