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> > adj;
vector<long long> arr;
pair<long long, long long> dfs(long long v, long long p = -1) {
  pair<long long, long long> res = {0, 1};
  for (long long x : adj[v]) {
    if (x == p) continue;
    pair<long long, long long> temp = dfs(x, v);
    res.first =
        ((((temp.first + temp.second) % 1000000007) * res.first % 1000000007) +
         (res.second * temp.first % 1000000007)) %
        1000000007;
    res.second =
        res.second * ((temp.first + temp.second) % 1000000007) % 1000000007;
  }
  if (arr[v] == 1) {
    res.first = res.second;
    res.second = 0;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  adj.resize(n);
  arr.resize(n);
  for (long long i = 1; i < n; i++) {
    long long x;
    cin >> x;
    adj[i].push_back(x);
    adj[x].push_back(i);
  }
  for (long long &x : arr) cin >> x;
  pair<long long, long long> ans = dfs(0);
  cout << ans.first << "\n";
}
