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;
int N;
vector<int> graph[100010];
long long dp0[100010], dp1[100010];
int color[100010];
int main(void) {
  int i, j;
  cin >> N;
  for ((i) = 0; (i) < (int)(N - 1); (i)++) {
    int p;
    scanf("%d", &p);
    graph[p].push_back(i + 1);
  }
  for ((i) = 0; (i) < (int)(N); (i)++) scanf("%d", &color[i]);
  for (i = N - 1; i >= 0; i--) {
    long long zero = 0, one = 0;
    if (color[i] == 0)
      zero++;
    else
      one++;
    for ((j) = 0; (j) < (int)(graph[i].size()); (j)++) {
      int v = graph[i][j];
      long long zero2 = (zero * dp1[v] + zero * dp0[v]) % 1000000007ll;
      long long one2 =
          (one * dp1[v] + zero * dp1[v] + one * dp0[v]) % 1000000007ll;
      zero = zero2;
      one = one2;
    }
    dp0[i] = zero;
    dp1[i] = one;
  }
  cout << dp1[0] << endl;
}
