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 N = 1e5 + 3;
long long n, m, i, j, k, l, s, t, d, u, ii, jj, x, y, z, h[N], hh[N][2];
vector<long long> V[N];
void f(int x) {
  int i, u = V[x].size();
  for (i = 0; i < u; i++) {
    f(V[x][i]);
    hh[x][1] = (hh[x][1] * (hh[V[x][i]][1] + hh[V[x][i]][0]) +
                hh[x][0] * hh[V[x][i]][1]) %
               1000000007;
    hh[x][0] = hh[x][0] * (hh[V[x][i]][0] + hh[V[x][i]][1]) % 1000000007;
  }
}
int main() {
  cin >> n;
  for (i = 2; i <= n; i++) {
    scanf("%lld", &x);
    V[x + 1].push_back(i);
  }
  for (i = 1; i <= n; i++) {
    scanf("%lld", &x);
    hh[i][x] = 1;
  }
  f(1);
  cout << hh[1][1] << '\n';
}
