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.
The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), 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 $x, y$ ($1 \leq x, y \leq n$), 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.
In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times.
5 4
abaca
1 2
1 3
3 4
4 5
3
6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4
-1
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
4
1 1
f
1 1
-1
10 50
ebibwbjihv
1 10
1 2
5 4
1 8
9 7
5 6
1 8
8 7
2 6
5 4
1 9
3 2
8 3
5 6
5 9
2 4
2 7
3 9
1 2
1 7
1 10
3 7
1 8
3 10
8 6
1 7
10 6
1 6
5 8
1 5
2 10
3 9
5 8
8 3
3 7
5 2
1 10
1 4
5 3
3 2
1 2
5 8
10 4
2 10
8 2
1 9
1 8
1 2
3 4
1 8
2
13 37
dwpzcppjmhkmz
2 6
3 6
6 7
6 7
6 7
6 7
6 8
6 8
6 8
6 8
4 6
4 6
5 6
4 6
4 6
6 9
6 9
6 10
6 10
6 10
6 10
4 6
1 6
1 6
10 11
6 11
1 6
6 12
6 12
6 12
6 13
6 13
6 13
6 13
3 6
2 6
2 6
3
5 8
tetqw
2 1
4 4
5 5
5 2
4 5
1 5
1 5
1 1
-1
5 8
reeet
4 3
2 5
4 2
2 4
4 2
5 2
3 3
3 4
-1
class Node:
    def __init__(self,val):
        self.val=val 
        self.left=None
        self.right=None
class BST:
    def __init__(self):
        self.root=None
    def insert(self,val):
      
        if self.root is None:
            self.root=Node(val)
        else:
            self._insert(val,self.root)
    
    def _insert(self,val,curnode):
        if val<curnode.val:
            if curnode.left is None:
                curnode.left=Node(val)
            else:
                self._insert(val,curnode.left)
        elif val>curnode.val:
            if curnode.right is None:
                curnode.right=Node(val)
            else:
                self._insert(val,curnode.right)
        else:
            print("element is already present in tree")
    def _find(self,val):
        if self.root:
            is_found=self.find(val,self.root)
            if is_found:
                return True
            return False
        else:
            return None


    def find(self,val,curnode):
        if  val==curnode.val: return True
        elif val<curnode.val and curnode.left:  return self.find(val,curnode.left)
        elif val>curnode.val and curnode.right:return self.find(val,curnode.right)


    def inorder(self):
        if self.root:
            li=self.print_inorder(self.root,[])
            print(*li)
            for i in range(len(li)-1):
                if li[i+1]<li[i]:
                    return False
            return True

    def print_inorder(self,node,li):
        if node:
            self.print_inorder(node.left,li)
            li.append(node.val)
            self.print_inorder(node.right,li)
        return li 

bst=BST()
'''
bst.insert(5)
bst.insert(4)
bst.insert(3)
bst.insert(6)
bst.insert(7)
bst.insert(8)
bst.insert(9)
bst.root.left.left.left=Node(10)

print(bst._find(2))
print(bst._find(4))

'''
bst.insert(8)
bst.root.left=Node(3)
bst.root.left.left=Node(2)

bst.root.left.right=Node(25)
bst.root.right=Node(10)


print(bst.inorder())