461_B. Appleman and Tree
Appleman has a tree with n vertices. Some of the vertices (at least one) are colored black and other vertices are colored white.

Consider a set consisting of k (0 ≤ k < n) edges of Appleman's tree. If Appleman deletes these edges from the tree, then it will split into (k + 1) parts. Note, that each part will be a tree with colored vertices.

Now Appleman wonders, what is the number of sets splitting the tree in such a way that each resulting part will have exactly one black vertex? Find this number modulo 1000000007 (109 + 7).

Input

The first line contains an integer n (2 ≤ n ≤ 105) — the number of tree vertices. 

The second line contains the description of the tree: n - 1 integers p0, p1, ..., pn - 2 (0 ≤ pi ≤ i). Where pi means that there is an edge connecting vertex (i + 1) of the tree and vertex pi. Consider tree vertices are numbered from 0 to n - 1.

The third line contains the description of the colors of the vertices: n integers x0, x1, ..., xn - 1 (xi is either 0 or 1). If xi is equal to 1, vertex i is colored black. Otherwise, vertex i is colored white.

Output

Output a single integer — the number of ways to split the tree modulo 1000000007 (109 + 7).

Examples

Input

3
0 0
0 1 1


Output

2


Input

6
0 1 1 0 4
1 1 0 0 1 0


Output

1


Input

10
0 1 2 1 4 4 4 0 8
0 0 0 1 0 1 1 0 0 1


Output

27
```python
from collections import UserDict


class Tree(UserDict):
    def __init__(self, g):
        super().__init__()
        for name, value in enumerate(g, 1):
            self[value] = name

    def __setitem__(self, name, value):
        if name in self:
            if value is not None:
                self[name].add(value)
                self[value] = None
        else:
            if value is None:
                super().__setitem__(name, set())
            else:
                super().__setitem__(name, {value})
                self[value] = None


if __name__ == '__main__':
    n = int(input())

    tree = Tree(int(i) for i in input().split())
    colors = [int(i) for i in input().split()]
    t = [()] * n

    def dfs(v):
        stack = [v]
        visited = set()

        while stack:
            v = stack.pop()
            if v not in visited:
                visited.add(v)
                stack.append(v)
                stack.extend(tree[v])
            else:
                t[v] = (1, colors[v])
                for u in tree[v]:
                    t[v] = (
                        (t[v][0] * t[u][1] + t[v][0] * t[u][0] * (not colors[u])) % (10**9 + 7),
                        (t[v][1] * t[u][1] + t[v][0] * t[u][1] * (not colors[v])
                                           + t[v][1] * t[u][0] * (not colors[u])) % (10**9 + 7)
                    )

                
    dfs(0)

    print(t[0][1])



    




# Made By Mostafa_Khaled
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