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>
const long base = 97;
using namespace std;
long long n, a[1000005];
vector<long> G[1000005];
long long F[1000005][2];
void nhap() {
  cin >> n;
  long x;
  for (long i = 1; i < n; i++) {
    cin >> x;
    G[x].push_back(i);
  }
  for (long i = 0; i < n; i++) cin >> a[i];
}
void DFS(long v) {
  F[v][a[v]] = 1;
  for (long i = 0; i < G[v].size(); i++) {
    long u = G[v][i];
    DFS(u);
    F[v][1] =
        ((F[v][1] * F[u][0]) % 1000000007 + (F[v][0] * F[u][1]) % 1000000007 +
         (F[v][1] * F[u][1]) % 1000000007) %
        1000000007;
    F[v][0] =
        ((F[v][0] * F[u][1]) % 1000000007 + (F[v][0] * F[u][0]) % 1000000007) %
        1000000007;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  nhap();
  DFS(0);
  cout << F[0][1] << endl;
}
