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;
long long a[int(1e6 + 6)], i, m, ans, k, l, j, x, n, ma, mi;
string s;
long long d[int(1e6 + 6)][2];
vector<long long> v[int(1e6 + 6)];
void go(int u) {
  d[u][0] = 1;
  d[u][1] = 0;
  for (int i = 0; i < v[u].size(); i++) {
    go(v[u][i]);
    int X = v[u][i];
    d[u][1] *= d[X][0];
    d[u][1] %= int(1e9) + 7;
    d[u][1] += d[u][0] * d[X][1];
    d[u][1] %= int(1e9) + 7;
    d[u][0] *= d[X][0];
    d[u][0] %= int(1e9) + 7;
  }
  if (a[u] == 1)
    d[u][1] = d[u][0];
  else
    d[u][0] += d[u][1];
  d[u][1] %= int(1e9) + 7;
  d[u][0] %= int(1e9) + 7;
}
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> x;
    x++;
    v[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  go(1);
  cout << d[1][1] << endl;
}
