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 long long int N = 1e5 + 1;
const long long int mod = 1e9 + 7;
vector<vector<long long int>> graph(N);
vector<long long int> color(N);
vector<long long int> white(N, 0), black(N, 0);
void dfs1(long long int node, long long int parent) {
  white[node] = (!color[node]);
  black[node] = color[node];
  for (auto i : graph[node]) {
    if (i == parent) continue;
    dfs1(i, node);
    long long int zero = white[i];
    long long int one = black[i];
    black[node] = ((black[node] * zero) % mod + (black[node] * one) % mod +
                   (white[node] * one) % mod) %
                  mod;
    white[node] =
        ((white[node] * zero) % mod + (white[node] * one) % mod) % mod;
  }
}
int main() {
  long long int n;
  cin >> n;
  for (long long int i = 1; i < n; i++) {
    long long int a;
    cin >> a;
    graph[i].push_back(a);
    graph[a].push_back(i);
  }
  for (long long int i = 0; i < n; cin >> color[i++])
    ;
  dfs1(0, 0);
  cout << black[0] << endl;
  return 0;
}
