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 double PI = acos(-1);
const int N = 1e5 + 10;
const int MOD = 1e9 + 7;
long long in[N], ex[N], col[N];
vector<int> child[N];
long long fex(int r);
long long fin(int r) {
  if (in[r] != -1) return in[r];
  if (col[r]) {
    in[r] = 1LL;
    for (long long i = 0; i < (child[r].size()); i++)
      in[r] = in[r] * ((fin(child[r][i]) + fex(child[r][i])) % MOD) % MOD;
  } else {
    in[r] = 0;
    long long mul = 1LL;
    for (long long i = 0; i < (child[r].size()); i++) {
      long long param = (fin(child[r][i]) + fex(child[r][i])) % MOD;
      in[r] = (in[r] * param % MOD + mul * fin(child[r][i]) % MOD) % MOD;
      mul = mul * param % MOD;
    }
  }
  return in[r];
}
long long fex(int r) {
  if (ex[r] != -1) return ex[r];
  if (col[r]) return ex[r] = 0;
  ex[r] = 1LL;
  for (long long i = 0; i < (child[r].size()); i++)
    ex[r] = ex[r] * ((fin(child[r][i]) + fex(child[r][i])) % MOD) % MOD;
  return ex[r];
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  memset(col, 0, sizeof(col));
  memset(in, -1, sizeof(in));
  memset(ex, -1, sizeof(ex));
  for (long long i = 0; i < (n - 1); i++) {
    int p;
    cin >> p;
    child[p].push_back(i + 1);
  }
  for (long long i = 0; i < (n); i++) cin >> col[i];
  cout << fin(0) << endl;
}
