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 MOD = 1000000007;
const int N = 100000 + 10;
int n;
vector<int> adj[N];
int color[N];
unsigned long long dp[N][2];
unsigned long long pow_mod(unsigned long long x, int n,
                           unsigned long long mod) {
  unsigned long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
unsigned long long inv(unsigned long long x) {
  if (x == 0 || x == 1) return 1;
  return pow_mod(x, MOD - 2, MOD);
}
void dfs(int t, int p) {
  for (int i = 0; i < adj[t].size(); i++) {
    int x = adj[t][i];
    if (x != p) dfs(x, t);
  }
  if (color[t]) {
    dp[t][1] = 1;
    dp[t][0] = 0;
    if (adj[t].size() == 1 && adj[t][0] == p) return;
    for (int i = 0; i < adj[t].size(); i++) {
      int x = adj[t][i];
      if (x != p) {
        dp[t][1] *= (dp[x][1] + dp[x][0]);
        dp[t][1] %= MOD;
      }
    }
  } else {
    dp[t][0] = 1;
    dp[t][1] = 0;
    if (adj[t].size() == 1 && adj[t][0] == p) return;
    for (int i = 0; i < adj[t].size(); i++) {
      int x = adj[t][i];
      if (x != p) {
        dp[t][0] *= (dp[x][0] + dp[x][1]);
        dp[t][0] %= MOD;
      }
    }
    for (int i = 0; i < adj[t].size(); i++) {
      int x = adj[t][i];
      if (x != p && dp[x][1] > 0) {
        dp[t][1] += dp[t][0] * inv(dp[x][0] + dp[x][1]) % MOD * dp[x][1] % MOD;
        dp[t][1] %= MOD;
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i <= n; i++) adj[i].clear();
  int t;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d", &t);
    adj[i + 1].push_back(t);
    adj[t].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &color[i]);
  }
  dfs(0, -1);
  printf("%llu\n", dp[0][1]);
  return 0;
}
