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> adj[100005];
long long dp[100005][2];
int color[100005];
void dfs(int u) {
  dp[u][0] = 1;
  dp[u][1] = 0;
  for (int v : adj[u]) {
    dfs(v);
    dp[u][1] *= dp[v][0];
    dp[u][1] %= 1000000007LL;
    dp[u][1] += dp[u][0] * dp[v][1] % 1000000007LL;
    dp[u][1] %= 1000000007LL;
    dp[u][0] *= dp[v][0];
    dp[u][0] %= 1000000007LL;
  }
  if (color[u])
    dp[u][1] = dp[u][0];
  else {
    dp[u][0] += dp[u][1];
    dp[u][0] %= 1000000007LL;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int v = 1; v < n; v++) {
    int u;
    scanf("%d", &u);
    adj[u].push_back(v);
  }
  for (int i = 0; i < n; i++) scanf("%d", color + i);
  dfs(0);
  printf("%I64d\n", dp[0][1]);
  return 0;
}
