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>
int const mo = 1000000007;
int const maxn = 100007;
bool black[maxn];
int n;
long long f[maxn][2];
std::vector<std::vector<int> > tree;
void add_edge(int u, int v) {
  tree[u].push_back(v);
  tree[v].push_back(u);
}
void dp(int x, int father) {
  f[x][0] = 1;
  f[x][1] = 0;
  for (int i = 0; i < (int)tree[x].size(); i++) {
    int v = tree[x][i];
    if (father == v) continue;
    dp(v, x);
    f[x][1] = (f[x][1] * f[v][0]) % mo;
    f[x][1] = (f[x][1] + f[x][0] * f[v][1]) % mo;
    f[x][0] = (f[x][0] * f[v][0]) % mo;
  }
  if (black[x])
    f[x][1] = f[x][0];
  else
    f[x][0] = (f[x][0] + f[x][1]) % mo;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n;
  tree.resize(n + 1);
  for (int i = 0, x; i < n - 1; i++) {
    std::cin >> x;
    add_edge(i + 2, x + 1);
  }
  for (int i = 1; i <= n; i++) std::cin >> black[i];
  dp(1, -1);
  std::cout << f[1][1] << "\n";
}
