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;
struct node {
  long long end, next;
} g[100100 * 2];
long long fa[100100], tot, n, i, ok[100100], x, f[100100][2];
long long ksm(long long a, long long b) {
  long long tem = 1ll;
  while (b) {
    if (b & 1) tem = tem * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return tem;
}
void add(long long a, long long b) {
  g[++tot].end = b;
  g[tot].next = fa[a];
  fa[a] = tot;
}
void dfs(long long w, long long pre) {
  long long i, tem, uk, s;
  for (i = fa[w]; i; i = g[i].next)
    if (pre != g[i].end) dfs(g[i].end, w);
  if (ok[w]) {
    tem = 1ll;
    for (i = fa[w]; i; i = g[i].next)
      if (pre != g[i].end) tem = tem * f[g[i].end][0] % 1000000007;
    f[w][0] = f[w][1] = tem;
  } else {
    tem = 0;
    uk = 1;
    for (i = fa[w]; i; i = g[i].next)
      if (pre != g[i].end) uk = uk * f[g[i].end][0] % 1000000007;
    for (i = fa[w]; i; i = g[i].next)
      if (pre != g[i].end) {
        s = uk * ksm(f[g[i].end][0], 1000000007 - 2) % 1000000007 *
            f[g[i].end][1] % 1000000007;
        tem = (tem + s) % 1000000007;
      }
    f[w][1] = tem;
    f[w][0] = (f[w][1] + uk) % 1000000007;
  }
}
int main() {
  scanf("%I64d", &n);
  for (i = 1; i < n; i++)
    scanf("%I64d", &x), add(x + 1, i + 1), add(i + 1, x + 1);
  for (i = 1; i <= n; i++) scanf("%I64d", &ok[i]);
  dfs(1, 0);
  printf("%I64d\n", f[1][1]);
  return 0;
}
