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 mXn = 1e5 + 10, M = 1e9 + 7;
vector<long long> ad[mXn];
bool b[mXn], mark[mXn];
long long dp0[mXn], dp1[mXn];
void dfs(long long v) {
  if (b[v]) {
    dp1[v] = 1;
  } else {
    dp0[v] = 1;
  }
  mark[v] = true;
  for (long long i = 0; i < (long long)ad[v].size(); i++) {
    long long u = ad[v][i];
    if (!mark[u]) {
      dfs(u);
      dp1[v] = (dp1[v] * (dp1[u] + dp0[u]) + dp0[v] * dp1[u]) % M;
      dp0[v] = (dp0[v] * (dp1[u] + dp0[u])) % M;
    }
  }
}
int32_t main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long x;
    cin >> x;
    ad[x].push_back(i + 1);
    ad[i + 1].push_back(x);
  }
  for (long long i = 0; i < n; i++) {
    long long f;
    cin >> f;
    b[i] = f;
  }
  dfs(0);
  cout << dp1[0];
  return 0;
}
