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 = 1e9 + 7;
int n;
vector<int> x;
vector<vector<int>> g;
vector<vector<long long>> dp;
long long BinPow(long long x, int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) return BinPow(x * x % MOD, n / 2);
  return x * BinPow(x, n - 1) % MOD;
}
long long Rev(long long x) { return BinPow(x, MOD - 2); }
void DFS(int u) {
  for (int v : g[u]) DFS(v);
  if (x[u] == 0) {
    dp[u][0] = 1;
    for (int v : g[u]) {
      dp[u][0] *= (dp[v][0] + dp[v][1]) % MOD;
      dp[u][0] %= MOD;
    }
    long long s = 1;
    for (int v : g[u]) {
      s *= (dp[v][0] + dp[v][1]) % MOD;
      s %= MOD;
    }
    dp[u][1] = 0;
    for (int v : g[u]) {
      dp[u][1] += ((dp[v][2] * s) % MOD) * Rev(dp[v][0] + dp[v][1]) % MOD;
      dp[u][1] %= MOD;
    }
    dp[u][2] = 0;
    for (int v : g[u]) {
      dp[u][2] += ((dp[v][2] * s) % MOD) * Rev(dp[v][0] + dp[v][1]) % MOD;
      dp[u][2] %= MOD;
    }
  }
  if (x[u] == 1) {
    dp[u][0] = 0;
    dp[u][1] = 1;
    for (int v : g[u]) {
      dp[u][1] *= (dp[v][0] + dp[v][1]) % MOD;
      dp[u][1] %= MOD;
    }
    dp[u][2] = 1;
    for (int v : g[u]) {
      dp[u][2] *= (dp[v][0] + dp[v][1]) % MOD;
      dp[u][2] %= MOD;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  g.resize(n);
  for (int i = 1; i < n; ++i) {
    int p;
    cin >> p;
    g[p].push_back(i);
  }
  x.resize(n);
  for (int i = 0; i < n; ++i) cin >> x[i];
  dp.resize(n, vector<long long>(3, 0));
  DFS(0);
  cout << dp[0][1] << endl;
  return 0;
}
