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;
vector<vector<long long int>> adj(100005);
vector<long long int> color(100005);
vector<vector<long long int>> dp(100005, vector<long long int>(2, 0));
long long int power(long long int x) {
  long long int res = 1;
  x %= 1000000007ll;
  long long int y = 1000000007ll - 2;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % 1000000007ll;
    }
    x = (x * x) % 1000000007ll;
    y = y >> 1;
  }
  return res;
}
void dfs(long long int v, long long int p) {
  long long int total = 1;
  for (auto child : adj[v]) {
    if (child == p) {
      continue;
    }
    dfs(child, v);
    total =
        (total * ((dp[child][0] + dp[child][1]) % 1000000007ll)) % 1000000007ll;
  }
  if (color[v] == 1) {
    dp[v][0] = 0;
    dp[v][1] = total;
  }
  if (color[v] == 0) {
    dp[v][0] = total;
    long long int sum = 0;
    for (auto child : adj[v]) {
      if (child == p) {
        continue;
      }
      sum = (sum +
             ((total * power((dp[child][0] + dp[child][1]) % 1000000007ll)) %
              1000000007ll * dp[child][1]) %
                 1000000007ll) %
            1000000007ll;
    }
    dp[v][1] = sum;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  long long int v;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> v;
    adj[v].push_back(i + 1);
    adj[i + 1].push_back(v);
  }
  for (long long int i = 0; i < n; i++) {
    cin >> color[i];
  }
  dfs(0, -1);
  cout << dp[0][1];
}
