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<long long> G[100010], ter;
long long n, root, c[100010], par[100010], dt[100010][2], ans;
void dfs(int v, int pv) {
  dt[v][0] = 1;
  dt[v][1] = 0;
  for (int w : G[v])
    if (w != pv) {
      dfs(w, v);
      dt[v][1] *= dt[w][0];
      dt[v][1] += dt[v][0] * dt[w][1];
      dt[v][0] *= dt[w][0];
      dt[v][1] %= 1000000007, dt[v][0] %= 1000000007;
    }
  if (c[v] == 1)
    dt[v][1] = dt[v][0];
  else
    dt[v][0] += dt[v][1];
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1, t; i < n; i++) {
    cin >> t;
    G[i].push_back(t), G[t].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> c[i];
  dfs(0, 0);
  cout << dt[0][1] << endl;
  return 0;
}
