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;
const int mod = 1000000007;
int ver[100005];
vector<int> ed[100005];
int ans = 1;
int inv(int b) {
  return (b == 1) ? 1 : 1ll * inv(mod % b) * (mod - mod / b) % mod;
}
void go(int wh, int p, int& adot, int& non) {
  int i1;
  int dottmp, nontmp;
  vector<int> vdot, vnon;
  non = 1;
  for (i1 = 0; i1 < ed[wh].size(); i1++) {
    if (ed[wh][i1] != p) {
      go(ed[wh][i1], wh, dottmp, nontmp);
      vdot.push_back(dottmp);
      vnon.push_back(nontmp);
      non = 1ll * non * nontmp % mod;
    }
  }
  adot = 0;
  for (i1 = vdot.size() - 1; i1 >= 0; i1--) {
    adot = (adot + 1ll * non * vdot[i1] % mod * inv(vnon[i1])) % mod;
  }
  if (ver[wh] == 1) {
    ans = 1ll * ans * non % mod;
    adot = 1;
    non = 1;
  } else {
    non = (non + adot) % mod;
  }
}
int main() {
  int i1;
  int n;
  int u;
  int dottmp, nontmp;
  scanf("%d", &n);
  for (i1 = 1; i1 < n; i1++) {
    scanf("%d", &u);
    ed[u].push_back(i1);
    ed[i1].push_back(u);
  }
  for (i1 = 0; i1 < n; i1++) {
    scanf("%d", &ver[i1]);
  }
  go(n - 1, -1, dottmp, nontmp);
  ans = ans * dottmp % mod;
  printf("%d\n", ans);
  return 0;
}
