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;
int n, u, v, x[int(1e5 + 10)];
long long int dp0[int(1e5 + 10)], dp1[int(1e5 + 10)];
vector<int> g[int(1e5 + 10)];
long long int power(long long int x, long long int n) {
  if (n == 0ll) return 1ll;
  long long int y = power(x, n / 2ll);
  y = (y * y) % int(1e9 + 7);
  if (n % 2ll) {
    return (x * y) % int(1e9 + 7);
  }
  return y;
}
void dfs(int s, int p) {
  dp0[s] = 1ll;
  dp1[s] = 0ll;
  long long int tmp;
  for (int i = 0; i < g[s].size(); i++) {
    if (g[s][i] != p) {
      dfs(g[s][i], s);
      dp0[s] = (dp0[s] * (dp0[g[s][i]] + dp1[g[s][i]]) % int(1e9 + 7)) %
               int(1e9 + 7);
    }
  }
  for (int i = 0; i < g[s].size(); i++) {
    if (g[s][i] != p) {
      tmp = (dp1[g[s][i]] * dp0[s]) % int(1e9 + 7);
      tmp = (tmp * power((dp0[g[s][i]] + dp1[g[s][i]]) % int(1e9 + 7),
                         int(1e9 + 7) - 2)) %
            int(1e9 + 7);
      dp1[s] = (dp1[s] + tmp) % int(1e9 + 7);
    }
  }
  if (x[s] == 1) {
    swap(dp0[s], dp1[s]);
    dp0[s] = 0;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(10);
  cin >> n;
  for (int i = 0; i + 1 < n; i++) {
    cin >> u;
    v = i + 1;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; i++) {
    cin >> x[i];
  }
  dfs(0, -1);
  cout << dp1[0] << endl;
  return 0;
}
