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>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, TT<T...>& c) {
  out << "{ ";
  for (auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const long long int MOD = 1e9 + 7, LIM = 1e5 + 5;
const long double EPS = 1e-9;
int n;
vector<int> adj[LIM];
bool visited[LIM];
long long int dp[LIM][2];
int x[LIM];
long long int dfs(int v) {
  dp[v][0] = 1;
  dp[v][1] = 0;
  for (auto& r : adj[v]) {
    if (visited[r]) continue;
    visited[r] = 1;
    dfs(r);
    dp[v][1] *= dp[r][0];
    dp[v][1] += (dp[v][0] * dp[r][1]);
    dp[v][1] %= MOD;
    dp[v][0] *= dp[r][0];
    dp[v][0] %= MOD;
  }
  if (x[v])
    dp[v][1] = dp[v][0];
  else
    dp[v][0] += dp[v][1];
  if (dp[v][0] > MOD) dp[v][0] -= MOD;
  return dp[v][1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x;
    cin >> x;
    adj[i + 1].push_back(x);
    adj[x].push_back(i + 1);
  }
  visited[0] = 1;
  for (int i = 0; i < n; i++) cin >> x[i];
  cout << dfs(0) << endl;
  return 0;
}
