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;
long long n, tot, e[500010], nt[500010], hd[500010], f[100010][2], p = 1e9 + 7,
                                                                   a[100010];
void build(long long x, long long y) {
  tot++;
  e[tot] = y;
  nt[tot] = hd[x];
  hd[x] = tot;
}
void dfs(long long x) {
  long long i;
  f[x][a[x]] = 1;
  for (i = hd[x]; i; i = nt[i]) {
    dfs(e[i]);
    f[x][1] =
        (f[x][1] * (f[e[i]][0] + f[e[i]][1]) % p + f[x][0] * f[e[i]][1] % p) %
        p;
    f[x][0] = f[x][0] * (f[e[i]][0] + f[e[i]][1]) % p;
  }
}
int main() {
  long long i, x;
  scanf("%lld", &n);
  for (i = 2; i <= n; i++) {
    scanf("%lld", &x);
    x++;
    build(x, i);
  }
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  dfs(1);
  printf("%lld", f[1][1]);
}
