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;
int parent[100100] = {0};
vector<int> child[100100];
int color[100100] = {0};
long long dyn_black[100100] = {0};
long long dyn_white[100100] = {0};
int run(int x, int par) {
  if (color[x] == 0) {
    dyn_white[x] = 1;
    dyn_black[x] = 0;
  } else {
    dyn_white[x] = 0;
    dyn_black[x] = 1;
  }
  for (int i = 0; i < child[x].size(); i++) run(child[x][i], x);
  if (par == -1) return dyn_black[x];
  long long parBlack = dyn_black[par];
  long long parWhite = dyn_white[par];
  dyn_black[par] = parWhite * dyn_black[x] + parBlack * dyn_white[x];
  dyn_white[par] = parWhite * dyn_white[x];
  dyn_white[par] += parWhite * dyn_black[x];
  dyn_black[par] += parBlack * dyn_black[x];
  dyn_black[par] %= 1000000007;
  dyn_white[par] %= 1000000007;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d", &parent[i]);
    child[parent[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) scanf("%d", &color[i]);
  printf("%lld", run(0, -1));
}
