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 int modul = 1000000007;
const int nmax = 100100;
const double e = 1e-8;
const double pi = acos(-1);
int n, m, stest, c[nmax], F[nmax][2], G[nmax];
vector<int> head[nmax];
void DFS(int u) {
  F[u][0] = (c[u] == 0);
  for (int i = 0; i <= (int)head[u].size() - 1; i++) {
    int v = head[u][i];
    DFS(v);
    F[u][0] = ((long long)F[u][0] * (F[v][0] + F[v][1])) % modul;
  }
  G[(int)head[u].size()] = 1;
  for (int i = (int)head[u].size() - 1; i >= 0; i--) {
    int v = head[u][i];
    G[i] = ((long long)G[i + 1] * (F[v][0] + F[v][1])) % modul;
  }
  F[u][1] = (c[u] == 1);
  int cur = 1;
  for (int i = 0; i <= (int)head[u].size() - 1; i++) {
    int v = head[u][i];
    if (c[u] == 1) {
      F[u][1] = ((long long)F[u][1] * (F[v][0] + F[v][1])) % modul;
    } else {
      F[u][1] =
          (F[u][1] + (((long long)cur * G[i + 1]) % modul) * F[v][1]) % modul;
      cur = ((long long)cur * (F[v][0] + F[v][1])) % modul;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int j;
    cin >> j;
    head[j].push_back(i);
  }
  for (int i = 0; i <= n - 1; i++) cin >> c[i];
  DFS(0);
  cout << F[0][1];
  return 0;
}
