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;
template <int POS, class TUPLE>
void deploy(std::ostream &os, const TUPLE &tuple) {}
template <int POS, class TUPLE, class H, class... Ts>
void deploy(std::ostream &os, const TUPLE &t) {
  os << (POS == 0 ? "" : ", ") << get<POS>(t);
  deploy<POS + 1, TUPLE, Ts...>(os, t);
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::vector<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::set<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T, class K>
std::ostream &operator<<(std::ostream &os, std::map<T, K> &make_pair) {
  int remain = make_pair.size();
  os << "{";
  for (auto e : make_pair)
    os << "(" << e.first << " -> " << e.second << ")"
       << (--remain == 0 ? "}" : ", ");
  return os;
}
int n;
vector<int> edges[100100];
bool color[100100];
long long memo[100100][2];
void dfs(int now) {
  memo[now][0] = 1;
  memo[now][1] = 0;
  for (auto u : edges[now]) {
    dfs(u);
    memo[now][1] *= memo[u][0];
    memo[now][1] %= 1000000007LL;
    memo[now][1] += (memo[now][0] * memo[u][1]) % 1000000007LL;
    memo[now][1] %= 1000000007LL;
    memo[now][0] *= memo[u][0];
    memo[now][0] %= 1000000007LL;
  }
  if (color[now]) {
    memo[now][1] = memo[now][0];
  } else {
    memo[now][0] += memo[now][1];
    memo[now][0] %= 1000000007LL;
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < (n); i++) {
    int x;
    cin >> x;
    edges[x].push_back(i);
  }
  for (int i = 0; i < (n); i++) {
    cin >> color[i];
  }
  dfs(0);
  cout << memo[0][1] << "\n";
  return 0;
}
