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;
bool c[100000];
vector<int> v[100000];
long long a[2][100000];
int const md = 1000000007;
void dfs(int n) {
  if (c[n]) {
    a[1][n] = 1;
    a[0][n] = 0;
    for (int i = 0; i < v[n].size(); ++i) {
      int h = v[n][i];
      dfs(h);
      a[1][n] *= (a[0][h] + a[1][h]);
      a[1][n] %= md;
    }
  } else {
    a[1][n] = 0;
    a[0][n] = 1;
    for (int i = 0; i < v[n].size(); ++i) {
      int h = v[n][i];
      dfs(h);
      a[1][n] *= a[0][h] + a[1][h];
      a[1][n] %= md;
      a[1][n] += a[1][h] * a[0][n];
      a[1][n] %= md;
      a[0][n] *= a[0][h] + a[1][h];
      a[0][n] %= md;
    }
  }
}
int main() {
  int n, t;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d", &t);
    v[t].push_back(i);
  }
  for (int i = 0; i < n; ++i) scanf("%d", &c[i]);
  dfs(0);
  cout << a[1][0];
}
