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
#include <bits/stdc++.h>
enum { M = 1000000007 };
int main() {
  int n;
  scanf("%d", &n);
  std::vector<std::vector<int>> G(n);
  std::vector<int> C(n);
  std::vector<long long> S(n), B(n);
  for (int i = (1); i <= (n - 1); ++i) {
    int a;
    scanf("%d", &a);
    G[a].push_back(i);
  }
  for (int &c : C) scanf("%d", &c);
  long long res = 1;
  for (int i = (n - 1); i >= (0); --i) {
    if (C[i]) {
      for (int j : G[i]) res = res * S[j] % M;
      S[i] = 1;
      B[i] = 1;
    } else {
      int d = G[i].size();
      std::vector<long long> a(d + 1), b(d + 1);
      a[0] = b[0] = 1;
      for (int j = 0; j < (d); ++j) {
        a[j + 1] = a[j] * S[G[i][j]] % M;
        b[j + 1] = b[j] * S[G[i][d - j - 1]] % M;
      }
      for (int j = 0; j < (d); ++j)
        B[i] += a[j] * B[G[i][j]] % M * b[d - j - 1] % M;
      B[i] %= M;
      S[i] = (a.back() + B[i]) % M;
    }
  }
  res = res * B[0] % M;
  printf("%d\n", int(res));
  return 0;
}
