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 as dd, deque

input=raw_input
range=xrange
n,m = map(int,input().split())
s = [ord(c)-ord('a') for c in input()]
E = dd(list)
P = dd(list)
C = [0]*n
for i in range(n):
    E[i] = []
for _ in range(m):
    a,b = map(int,input().split())
    E[a-1].append(b-1)
    P[b-1].append(a-1)
    C[a-1] += 1

leafs = [v for v in E if len(E[v])==0]

used = [0]*n
DP = [[0]*27 for _ in range(n)]

for i in range(n):
    DP[i][s[i]] += 1

Q = deque()
for leaf in leafs:
    Q.append(leaf)

while Q:
    node = Q.pop()
    V = []
    for i in range(27):
        mx = 0
        for ch in E[node]:
            mx = max(mx,DP[ch][i])
        DP[node][i] += mx
    if used[leaf]: continue
    for pa in P[node]:
        C[pa] -= 1
        if C[pa] == 0:
            Q.append(pa)

if any(C):
    print(-1)
else:
    print(max(max(dp) for dp in DP))