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 double eps(1e-8);
const double pi(3.14159265358979);
const int N = 100100, mod = 1000000007;
int n;
int color[N] = {};
long long f[N][2] = {};
vector<int> v[N];
void init() {
  scanf("%d", &n);
  for (int i = 2, x = 0; i <= n; ++i) {
    scanf("%d", &x);
    v[x + 1].push_back(i);
  }
  for (int i = 1; i <= n; ++i) scanf("%d", &color[i]);
}
void dfs(int s) {
  f[s][color[s]] = 1;
  long long x1 = 0, x2 = 0;
  for (vector<int>::iterator it = v[s].begin(); it != v[s].end(); ++it) {
    dfs(*it);
    x1 = f[s][0] * f[*it][0];
    x2 = f[s][1] * f[*it][0] + f[s][0] * f[*it][1];
    f[s][0] = x1 % mod;
    f[s][1] = x2 % mod;
  }
  f[s][0] = (f[s][0] + f[s][1]) % mod;
}
void work() {
  dfs(1);
  cout << f[1][1] << endl;
}
int main() {
  init();
  work();
  return 0;
}
