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;
inline void OPEN(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
int n, p[100100], c[100100];
vector<int> child[100100];
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = power(a, b / 2LL);
  temp *= temp;
  temp %= 1000000007LL;
  if (b & 1) {
    temp *= a;
    temp %= 1000000007LL;
  }
  return temp;
}
long long inv(long long x) { return power(x, 1000000007LL - 2LL); }
pair<long long, long long> dfs(int v) {
  vector<pair<long long, long long> > t;
  int czero = 0;
  for (__typeof((child[v]).begin()) it = (child[v]).begin();
       it != (child[v]).end(); it++) {
    pair<long long, long long> temp = dfs(*it);
    if (temp.first != 0 || temp.second != 0) {
      t.push_back(temp);
      if ((temp.first + temp.second) % 1000000007LL == 0) czero++;
    }
  }
  if (t.empty()) {
    if (c[v] == 1) {
      return make_pair(0, 1);
    } else {
      return make_pair(0, 0);
    }
  }
  long long x = 1;
  long long xzero = 1;
  for (__typeof((t).begin()) it = (t).begin(); it != (t).end(); it++) {
    x *= (it->first + it->second);
    x %= 1000000007LL;
    if ((it->first + it->second) % 1000000007LL != 0) {
      xzero *= (it->first + it->second);
      xzero %= 1000000007LL;
    }
  }
  long long y = 0;
  for (__typeof((t).begin()) it = (t).begin(); it != (t).end(); it++) {
    long long ty = it->second;
    if (czero == 0) {
      ty *= x;
      ty %= 1000000007LL;
      ty *= inv(it->first + it->second);
      ty %= 1000000007LL;
    } else if (czero == 1) {
      if ((it->first + it->second) % 1000000007LL == 0) {
        ty *= xzero;
        ty %= 1000000007LL;
      } else {
        ty = 0;
      }
    } else {
      ty = 0;
    }
    y += ty;
    y %= 1000000007LL;
  }
  if (c[v] == 1) {
    y = x;
    x = 0;
  }
  return make_pair(x, y);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n - 1); ++i) {
    scanf("%d", &p[i + 1]);
    child[p[i + 1]].push_back(i + 1);
  }
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &c[i]);
  }
  pair<long long, long long> ans = dfs(0);
  cout << ans.second << endl;
  return 0;
}
