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;
const int N = (int)1e5 + 5;
const long long MOD = (long long)1e9 + 7;
vector<int> adj[N];
int n, x[N];
long long dp[N][2];
void dfs(int u) {
  dp[u][0] = 1LL;
  dp[u][1] = 0LL;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    dfs(v);
    dp[u][1] = (dp[u][1] % MOD * dp[v][0] % MOD) % MOD;
    dp[u][1] =
        (dp[u][1] % MOD + ((dp[u][0] % MOD * dp[v][1] % MOD) % MOD) % MOD) %
        MOD;
    dp[u][0] = (dp[u][0] % MOD * dp[v][0] % MOD) % MOD;
  }
  if (x[u] == 1)
    dp[u][1] = dp[u][0];
  else
    dp[u][0] = (dp[u][0] % MOD + dp[u][1] % MOD) % MOD;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int p;
    scanf("%d", &p);
    adj[p].push_back(i);
  }
  for (int i = 0; i < n; i++) scanf("%d", &x[i]);
  dfs(0);
  printf("%lld\n", dp[0][1] % MOD);
  return 0;
}
