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 long long MOD = 1000000007;
const int MAX = 100005;
int B[MAX], N;
long long DP[MAX][2];
vector<int> E[MAX];
void dfs(int n, int p) {
  DP[n][1] = B[n];
  DP[n][0] = 1 - B[n];
  for (int v : E[n])
    if (v != p) {
      long long old[2];
      old[0] = DP[n][0];
      old[1] = DP[n][1];
      DP[n][0] = DP[n][1] = 0;
      dfs(v, n);
      DP[n][0] += DP[v][0] * old[0];
      DP[n][0] %= MOD;
      DP[n][1] += DP[v][1] * old[0];
      DP[n][1] %= MOD;
      DP[n][1] += DP[v][0] * old[1];
      DP[n][1] %= MOD;
      DP[n][0] += DP[v][1] * old[0];
      DP[n][0] %= MOD;
      DP[n][1] += DP[v][1] * old[1];
      DP[n][1] %= MOD;
    }
}
int main() {
  cin >> N;
  for (int i = (2); i <= (N); i++) {
    int p;
    cin >> p;
    E[p + 1].push_back(i), E[i].push_back(p + 1);
  }
  for (int i = (1); i <= (N); i++) cin >> B[i];
  dfs(1, 0);
  cout << DP[1][1];
}
