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;
vector<int> graph[100000];
bool colour[100000];
long long dp[100000][2];
void dfs(int vertex, int parent) {
  int i, child;
  dp[vertex][1] = colour[vertex];
  dp[vertex][0] = 1 - colour[vertex];
  for (i = 0; i < graph[vertex].size(); i++) {
    child = graph[vertex][i];
    if (child == parent) continue;
    dfs(child, vertex);
    if (colour[vertex]) {
      dp[vertex][1] *= ((dp[child][0] + dp[child][1]) % 1000000007);
      dp[vertex][1] = (dp[vertex][1]) % 1000000007;
    } else {
      dp[vertex][1] *= ((dp[child][0] + dp[child][1]) % 1000000007);
      dp[vertex][1] = (dp[vertex][1]) % 1000000007;
      dp[vertex][1] += ((dp[vertex][0] * dp[child][1]) % 1000000007);
      dp[vertex][1] = (dp[vertex][1]) % 1000000007;
      dp[vertex][0] *= ((dp[child][0] + dp[child][1]) % 1000000007);
      dp[vertex][0] = (dp[vertex][0]) % 1000000007;
    }
  }
}
int main() {
  int n;
  int i;
  int vertex;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d", &vertex);
    graph[i].push_back(vertex);
    graph[vertex].push_back(i);
  }
  for (i = 0; i < n; i++) scanf("%d", &colour[i]);
  dfs(0, -1);
  printf("%I64d\n", dp[0][1]);
  return 0;
}
