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;
long long int power(long long int a, long long int b, long long int m) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long int modInverse(long long int num) {
  num = power(num, 1000000007 - 2, 1000000007);
  return num;
}
long long int dp[300009][2];
long long int c[300009];
vector<long long int> v[300009];
void dfs(long long int num, long long int par) {
  long long int tot = 1;
  for (auto it : v[num]) {
    if (it == par) continue;
    dfs(it, num);
    tot *= (dp[it][0] + dp[it][1]);
    tot %= 1000000007;
  }
  if (c[num]) {
    dp[num][1] = tot;
    dp[num][0] = 0;
  } else {
    dp[num][0] = tot;
    long long int ans = 0;
    for (auto it : v[num]) {
      if (it == par) continue;
      ans += ((dp[it][1] * (modInverse(dp[it][0] + dp[it][1]))) % 1000000007 *
              tot) %
             1000000007;
      ans %= 1000000007;
    }
    dp[num][1] = ans;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n - 1; i++) {
    long long int x;
    cin >> x;
    v[i + 1].push_back(x);
    v[x].push_back(i + 1);
  }
  for (long long int i = 0; i < n; i++) cin >> c[i];
  dfs(0, -1);
  cout << dp[0][1] << endl;
}
