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 int sz = 100005;
int n;
vector<int> T[sz];
vector<int> col;
const int MOD = 1000000007;
long long P[sz][2];
void dfs(int u, int prev) {
  if (col[u] == 1)
    P[u][1] = 1, P[u][0] = 0;
  else
    P[u][1] = 0, P[u][0] = 1;
  for (int i = 0; i < T[u].size(); i++) {
    int v = T[u][i];
    if (v == prev) continue;
    dfs(v, u);
    P[u][1] *= (P[v][0] + P[v][1]);
    P[u][1] %= MOD;
    P[u][1] += P[u][0] * P[v][1];
    P[u][1] %= MOD;
    P[u][0] = P[u][0] * (P[v][0] + P[v][1]);
    P[u][0] %= MOD;
  }
}
int main() {
  memset(P, 0, sizeof P);
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int x;
    scanf("%d", &x);
    T[i + 1].push_back(x);
    T[x].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    col.push_back(x);
  }
  dfs(0, -1);
  cout << P[0][1] << "\n";
  return 0;
}
