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 getpow(long long a, long long b) {
  long long ret = 1LL;
  long long inter = a % 1000000007;
  while (b) {
    if (b & 1) {
      ret *= inter;
      ret %= 1000000007;
    }
    inter *= inter;
    inter %= 1000000007;
    b = b >> 1;
  }
  return ret;
}
int n, p;
vector<int> edges[1000000];
int brr[1000000];
int vis[1000000] = {0};
void in() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> p;
    edges[p].push_back(i + 1);
    edges[i + 1].push_back(p);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &brr[i]);
  }
}
pair<long long, long long> dfs(int u) {
  vis[u] = 1;
  vector<pair<long long, long long> > ret;
  long long s = 1;
  for (int i = 0; i < edges[u].size(); i++) {
    int v = edges[u][i];
    if (vis[v] == 0) {
      ret.push_back(dfs(v));
    }
  }
  if (ret.size() == 0) {
    if (brr[u] == 1) {
      return make_pair(1, 0);
    } else {
      return make_pair(0, 1);
    }
  }
  for (int i = 0; i < ret.size(); i++) {
    s *= (ret[i].first + ret[i].second);
    s %= 1000000007;
  }
  if (brr[u] == 1) {
    return make_pair(s, 0);
  }
  long long a = 0, b;
  for (int i = 0; i < ret.size(); i++) {
    b = getpow(ret[i].first + ret[i].second, 1000000007 - 2);
    b *= s;
    b %= 1000000007;
    b *= ret[i].first;
    b %= 1000000007;
    a += b;
    a %= 1000000007;
  }
  return make_pair(a, s);
}
void solve() { cout << (dfs(0).first) % 1000000007 << endl; }
int main() {
  in();
  solve();
}
