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 long long maxinput = 1e5 + 4;
long long n, col[maxinput], parent[maxinput];
vector<long long> adj[maxinput];
long long dp[maxinput][2];
long long expo(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent != 0) {
    if ((exponent & 1) == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base = base * base;
    base %= mod;
    exponent >>= 1;
  }
  return ans % mod;
}
long long inv(long long x) { return expo(x, 1000000007 - 2, 1000000007); }
void dfs(long long u) {
  long long v;
  dp[u][0] = 0LL;
  dp[u][1] = 0LL;
  long long prod = 1LL;
  long long temp_sum = 0LL;
  for (long long j = 0; j < adj[u].size(); j++) {
    v = adj[u][j];
    if (v != parent[u]) {
      parent[v] = u;
      dfs(v);
      prod *= (dp[v][0] + dp[v][1]);
      prod %= 1000000007;
      temp_sum += (dp[v][1] * inv(dp[v][0] + dp[v][1]));
      temp_sum %= 1000000007;
    }
  }
  temp_sum = (temp_sum * prod) % 1000000007;
  if ((adj[u].size() == 1) && adj[u][0] == parent[u]) {
    if (col[u] == 1)
      dp[u][1] = 1LL;
    else
      dp[u][0] = 1LL;
    return;
  }
  if (col[u] == 0) {
    dp[u][0] = prod;
    dp[u][0] %= 1000000007;
    dp[u][1] = temp_sum;
    dp[u][1] %= 1000000007;
  } else {
    dp[u][0] = 0;
    dp[u][1] = prod;
    dp[u][1] %= 1000000007;
  }
}
int main() {
  cin >> n;
  long long u, v;
  for (long long i = 1; i < n; i++) {
    cin >> u;
    adj[u].push_back(i);
    adj[i].push_back(u);
  }
  for (long long i = 0; i < n; i++) cin >> col[i];
  parent[0] = -1;
  dfs(0);
  cout << dp[0][1] << endl;
}
