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;
int countBits(long long mask) {
  int res = 0;
  while (mask) mask &= (mask - 1), ++res;
  return res;
}
string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
long long toNumber(string s) {
  stringstream ss;
  long long n;
  ss << s;
  ss >> n;
  return n;
}
long long dp[100010][2];
bool black[100010];
vector<vector<int> > adjList;
long long mod_inverse(long long n, long long p) {
  if (!p) return 1;
  if (p == 1) return n;
  long long res = mod_inverse(n, p / 2);
  res = (res * res) % 1000000007;
  if (p & 1) res = (res * n) % 1000000007;
  return res;
}
long long rec(int node, bool has_black_ancestor, int parent) {
  if (adjList[node].size() == 1 && parent != -1)
    return black[node] || has_black_ancestor;
  long long &res = dp[node][has_black_ancestor];
  if (res != -1) return res;
  res = 0;
  long long all_ok = 1;
  for (int i = (int)(0), _m = (int)(adjList[node].size()); i < _m; ++i) {
    int child = adjList[node][i];
    if (child == parent) continue;
    all_ok = (all_ok * rec(child, 1, node)) % 1000000007;
  }
  if (black[node] || has_black_ancestor) res = all_ok;
  if (!black[node]) {
    for (int i = (int)(0), _m = (int)(adjList[node].size()); i < _m; ++i) {
      int child = adjList[node][i];
      if (child == parent) continue;
      long long all_but_this =
          (all_ok * mod_inverse(rec(child, 1, node), 1000000007 - 2)) %
          1000000007;
      res += (all_but_this * rec(child, 0, node)) % 1000000007;
      res %= 1000000007;
    }
  }
  return res;
}
int main() {
  int n, p;
  while (cin >> n) {
    adjList.clear();
    adjList.resize(n);
    for (int i = (int)(1), _m = (int)(n); i < _m; ++i) {
      cin >> p;
      adjList[p].push_back(i);
      adjList[i].push_back(p);
    }
    for (int i = (int)(0), _m = (int)(n); i < _m; ++i) cin >> black[i];
    memset(dp, -1, sizeof dp);
    cout << rec(0, 0, -1) << endl;
  }
}
