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, son[1000005], cnt = 0;
long long f[1000005][2];
int read() {
  int x = 0, f = 1;
  char h = getchar();
  for (; !isdigit(h); h = getchar())
    if (h == '-') f = -1;
  for (; isdigit(h); h = getchar()) x = (x << 3) + (x << 1) + h - 48;
  return x * f;
}
int main() {
  n = read();
  memset(f, 0, sizeof(f));
  for (int i = 1; i < n; i++) son[i] = read();
  for (int i = 0; i < n; i++) f[i][read()] = 1;
  for (int i = n - 1; i; i--)
    (f[son[i]][1] = f[son[i]][0] * f[i][1] % 1000000007 +
                    f[son[i]][1] * (f[i][0] + f[i][1]) % 1000000007) %=
        1000000007,
        (f[son[i]][0] = f[son[i]][0] * (f[i][0] + f[i][1])) %= 1000000007;
  cout << f[0][1];
  return 0;
}
