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>
using namespace std;
const int mod = 1000000007;
struct Edge {
  int v;
  Edge *next;
};
int n, p[100000], x[100000], f[100000], g[100000];
vector<int> ed[100000];
int ExEuler(int a, int b, int c) {
  c = c < 0 ? c + (-(c + 1) / b + 1) * b : c % b;
  return a ? ((long long)ExEuler(b % a, a, -c) * b + c) / a : 0;
}
int Div(int a, int b) { return ExEuler(b, mod, a); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d", &p[i]), ed[p[i]].push_back(i);
  for (int i = 0; i < n; i++) scanf("%d", &x[i]);
  for (int i = n - 1; i >= 0; i--) {
    if (x[i]) {
      f[i] = 1;
      for (int j = 0; j < ed[i].size(); j++)
        f[i] = (long long)f[i] * (g[ed[i][j]] + f[ed[i][j]]) % mod;
    } else {
      g[i] = 1;
      for (int j = 0; j < ed[i].size(); j++)
        g[i] = (long long)g[i] * (g[ed[i][j]] + f[ed[i][j]]) % mod;
      for (int j = 0; j < ed[i].size(); j++)
        f[i] = ((long long)Div(g[i], g[ed[i][j]] + f[ed[i][j]]) * f[ed[i][j]] +
                f[i]) %
               mod;
    }
  }
  printf("%d\n", f[0]);
  getchar();
  getchar();
  return 0;
}
