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 int oo = 2000000009;
const double eps = 1e-9;
list<int> adj[100005];
int label[100005];
long long dp[100005][2];
const long long MOD = 1000000007;
void numWays(int v, int p) {
  dp[v][0] = 1 - label[v];
  dp[v][1] = label[v];
  for (typeof(adj[v].begin()) u = adj[v].begin(); u != adj[v].end(); u++)
    if (*u != p) {
      long long cum[2];
      numWays(*u, v);
      cum[0] = dp[v][0];
      cum[1] = dp[v][1];
      dp[v][0] = dp[v][1] = 0;
      dp[v][1] += dp[*u][0] * cum[1];
      dp[v][1] %= MOD;
      dp[v][1] += dp[*u][1] * cum[0];
      dp[v][1] %= MOD;
      dp[v][0] += dp[*u][0] * cum[0];
      dp[v][0] %= MOD;
      dp[v][1] += dp[*u][1] * cum[1];
      dp[v][1] %= MOD;
      dp[v][0] += dp[*u][1] * cum[0];
      dp[v][0] %= MOD;
    }
}
int main() {
  int n, temp;
  scanf("%d", &n);
  for (int i = 0; i < (n - 1); ++i) {
    scanf("%d", &temp);
    adj[i + 1].push_back(temp);
    adj[temp].push_back(i + 1);
  }
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &label[i]);
  }
  numWays(0, -1);
  cout << dp[0][1] << endl;
  return 0;
}
