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>
#pragma GCC optimize("Ofast")
#pragma GCC target( \
    "avx,avx2,fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long MOD = 1e9 + 7;
const int N = 200000;
vector<long long> gr[N];
long long sz[N];
long long st(long long t, long long id, long long mod) {
  long long res = 1, a = t, n = id;
  while (n) {
    if (n & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    n >>= 1;
  }
  return res % mod;
}
long long del(long long a, long long b) {
  return (a * st(b, MOD - 2, MOD)) % MOD;
}
pair<long long, long long> dfs(long long v, long long pr = -1) {
  pair<long long, long long> ans = {0, 1};
  vector<pair<long long, long long> > gg;
  for (auto i : gr[v]) {
    if (i != pr) {
      auto g = dfs(i, v);
      gg.push_back(g);
      ans.second = ans.second * (g.second + g.first) % MOD;
    }
  }
  for (auto g : gg) {
    ans.first += del(ans.second, (g.second + g.first)) * g.first % MOD;
    ans.first %= MOD;
  }
  if (sz[v]) {
    ans.first = ans.second;
    ans.second = 0;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0),
      cout << fixed << setprecision(20);
  long long n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    long long u, v = i + 1;
    cin >> u;
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  for (int i = 0; i < n; i++) {
    cin >> sz[i];
  }
  auto mem = dfs(0, 0);
  cout << mem.first;
}
