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 MX = 2e5;
vector<long long> adj[MX];
const long long MOD = 1e9 + 7;
long long mult(long long x, long long y) { return (x * y) % MOD; }
long long add(long long x, long long y) { return (x + y) % MOD; }
long long fastPow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1 == 1) {
      ans = mult(ans, a);
    }
    a = mult(a, a);
    b >>= 1;
  }
  return ans;
}
long long modInverse(long long x) { return fastPow(x, MOD - 2); }
long long dp[MX][2];
bool used[MX][2];
bool isBlack[MX];
long long cnt[MX];
long long count(long long u, bool black) {
  if (used[u][black]) return dp[u][black];
  if (isBlack[u] && black) return 0;
  used[u][black] = true;
  long long ans = 1;
  if (isBlack[u] || black) {
    for (long long v : adj[u]) {
      ans = mult(ans, add(count(v, true), count(v, false)));
    }
  } else {
    if (cnt[u] == 0) {
      ans = 0;
    } else {
      long long prod = 1;
      for (long long v : adj[u]) {
        long long x = add(count(v, false), count(v, true));
        prod = mult(prod, x);
      }
      ans = 0;
      for (long long i = 0; i < adj[u].size(); i++) {
        long long v = adj[u][i];
        long long partial = mult(prod, count(v, false));
        long long x = add(count(v, false), count(v, true));
        partial = mult(partial, modInverse(x));
        ans = add(ans, partial);
      }
    }
  }
  return dp[u][black] = ans;
}
long long dfs(long long u) {
  if (isBlack[u]) cnt[u]++;
  for (long long v : adj[u]) {
    cnt[u] += dfs(v);
  }
  return cnt[u];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long p;
    cin >> p;
    adj[p].push_back(i);
  }
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    isBlack[i] = x;
  }
  dfs(0);
  cout << count(0, false) << "\n";
  return 0;
}
