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;
#pragma comment(linker, "/STACK:33554432")
const int MOD = 1e9 + 7;
int N, K, A[100005], par[100005];
int D[100005][2];
vector<int> con[100005];
int inv(int n) {
  int ret = 1, v = n;
  for (int i = 0; i < 30; i++, v = (long long)v * v % MOD)
    if (((MOD - 2) >> i) & 1) {
      ret = (long long)ret * v % MOD;
    }
  return ret;
}
void dfs(int n) {
  int i, k;
  D[n][0] = 1;
  for (i = ((int)(con[n]).size()); i--;) {
    k = con[n][i];
    dfs(k);
    D[n][0] = (long long)D[n][0] * D[k][0] % MOD;
  }
  if (A[n])
    D[n][1] = D[n][0];
  else {
    for (i = ((int)(con[n]).size()); i--;) {
      k = con[n][i];
      int d = (long long)D[n][0] * inv(D[k][0]) % MOD;
      D[n][1] = (D[n][1] + (long long)D[k][1] * d % MOD) % MOD;
    }
    D[n][0] = (D[n][0] + (D[n][1])) % (MOD);
  }
}
int main() {
  int i, j, k;
  scanf("%d", &N);
  for (i = 2; i <= N; i++)
    scanf("%d", par + i), ++par[i], con[par[i]].push_back(i);
  for (i = 1; i <= N; i++) scanf("%d", A + i), K += A[i];
  dfs(1);
  printf("%d\n", D[1][1]);
}
