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 N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int n, i, x;
int dp[N][2], A[N];
vector<int> V[N];
int f(int x, int k) {
  int &r = dp[x][k], i;
  if (r != -1) return r;
  if (A[x]) {
    r = 1;
    for (i = 0; i < V[x].size(); i++) r = (long long)r * f(V[x][i], 0) % mod;
    return r;
  }
  if (!V[x].size()) return r = !k;
  vector<int> pre, suf;
  pre.resize(V[x].size() + 5);
  suf.resize(V[x].size() + 5);
  pre[0] = suf[V[x].size()] = 1;
  for (i = 0; i < V[x].size(); i++)
    pre[i + 1] = (long long)pre[i] * f(V[x][i], 0) % mod;
  for (i = V[x].size() - 1; i >= 0; i--)
    suf[i] = (long long)suf[i + 1] * f(V[x][i], 0) % mod;
  r = 0;
  for (i = 0; i < V[x].size(); i++)
    r = (r + (long long)f(V[x][i], 1) * pre[i] % mod * suf[i + 1] % mod) % mod;
  if (!k) r = ((long long)r + suf[0]) % mod;
  return r;
}
int main() {
  scanf("%d", &n);
  for (i = 2; i <= n; i++) {
    scanf("%d", &x);
    V[x + 1].push_back(i);
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    A[i] = x;
  }
  memset(dp, -1, sizeof dp);
  cout << f(1, 1) << endl;
}
