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;
int x[100000];
vector<int> adj[100000];
void add(int &a, int b) { a = (a + b) % 1000000007; }
int mul(int a, int b) { return (long long)a * b % 1000000007; }
pair<int, int> dfs(int u) {
  pair<int, int> ret = {1, 0};
  for (int v : adj[u]) {
    pair<int, int> p = dfs(v);
    ret.second = mul(ret.second, p.first);
    add(ret.second, mul(ret.first, p.second));
    ret.first = mul(ret.first, p.first);
  }
  if (x[u])
    ret.second = ret.first;
  else
    add(ret.first, ret.second);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int p;
    cin >> p;
    adj[p].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> x[i];
  cout << dfs(0).second << "\n";
  return 0;
}
