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>
const int N = 100005, mo = 1000000007;
int f[N][2], a[N], q[N], fi[N], v[N], tot = 0;
struct Edge {
  int to, ne;
} e[N << 1];
void add(int x, int y) {
  e[++tot].to = y;
  e[tot].ne = fi[x];
  fi[x] = tot;
}
void dfs(int x, int fa) {
  f[x][0] = 1;
  for (int i = fi[x]; i; i = e[i].ne)
    if (e[i].to != fa) {
      int y = e[i].to;
      dfs(y, x);
      f[x][0] = 1ll * f[x][0] * f[y][0] % mo;
    }
  if (v[x]) {
    f[x][1] = f[x][0];
    return;
  }
  int t = 0, z = 1;
  a[0] = 1;
  for (int i = fi[x]; i; i = e[i].ne)
    if (e[i].to != fa) {
      q[++t] = e[i].to;
      a[t] = 1ll * a[t - 1] * f[q[t]][0] % mo;
    }
  for (int i = t; i >= 1; i--) {
    f[x][1] = (f[x][1] + 1ll * z * a[i - 1] % mo * f[q[i]][1]) % mo;
    z = 1ll * z * f[q[i]][0] % mo;
  }
  (f[x][0] += f[x][1]) %= mo;
}
int main() {
  int n;
  scanf("%d\n", &n);
  for (int i = 1; i < n; i++) {
    int x;
    scanf("%d", &x);
    add(x, i);
    add(i, x);
  }
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  dfs(0, -1);
  printf("%d\n", f[0][1]);
}
