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.
maxi = 300005
n, m = map(int, input().split())
s = " "+input()
ed,dp = [], []

for _ in range(n+1):
    ed.append([])
    dp.append([0]*26)
q,deg = [0]*maxi,[0]*(n+1)

for i in range(m):
    x, y = map(int, input().split())
    ed[x].append(y)
    deg[y] += 1
    
fr,re = 0,0
def funct():
    re = fr = 0
    for i in range(1, n+1):
        if(deg[i] == 0):
            q[re] = i
            re += 1
            dp[i][ord(s[i])-97] = 1
    count = 0
    
    while(fr != re):
        u = q[fr]
        fr += 1
        count += 1
        for i in range(len(ed[u])):
            v = ed[u][i]
            for j in range(26):
                x = dp[u][j]
                if(j == ord(s[v])-ord('a')):
                    x += 1
                dp[v][j] = max(x, dp[v][j])
            deg[v] -= 1
            if(deg[v] == 0):
                q[re] = v
                re += 1
                
    ans = -1
    for i in range(1, n+1):
        for j in range(26):
            ans = max(ans, dp[i][j])
            
    return count == n, ans
 
 
ans1 = funct()
if(ans1[0]):
    print(ans1[1])
else:
    print(-1)

                    
