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 n, i, p[100010], a, fa;
bool x[100010];
long long ans[100010][2], an1, an0;
const long long kkk = 1000000007;
priority_queue<long long> q1;
int main() {
  scanf("%d", &n);
  for (i = 1; i < n; i++) scanf("%d", &p[i]);
  for (i = 0; i < n; i++) {
    scanf("%d", &a);
    if (a) {
      x[i] = 1;
      ans[i][1] = 1;
    } else {
      x[i] = 0;
      ans[i][0] = 1;
    }
  }
  for (i = n - 1; i > 0; i--) {
    fa = p[i];
    an0 = (ans[fa][0] * ans[i][0] % kkk + ans[fa][0] * ans[i][1] % kkk) % kkk;
    an1 = (ans[fa][0] * ans[i][1] % kkk + ans[fa][1] * ans[i][1] % kkk +
           ans[fa][1] * ans[i][0] % kkk) %
          kkk;
    ans[fa][0] = an0;
    ans[fa][1] = an1;
  }
  printf("%I64d\n", ans[0][1]);
  return 0;
}
