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;
const int Maxn = 1e5 + 1e3;
const long long Mod = 1e9 + 7;
vector<int> al[Maxn];
int bad[Maxn], dp[Maxn][2];
long long pw(long long a, long long b) {
  if (!b) return 1;
  return (pw((a * a) % Mod, b / 2) * (b % 2 ? a : 1)) % Mod;
}
long long dfs(int v, int p = -1, int state = 1) {
  if (dp[v][state] != -1) return dp[v][state];
  if (bad[v]) {
    if (!state) return 0;
    long long ret = 1;
    for (int i = (0); i < ((int)al[v].size()); ++i) {
      int u = al[v][i];
      if (u == p) continue;
      ret *= (dfs(u, v, 0) + dfs(u, v, 1));
      ret %= Mod;
    }
    return dp[v][state] = ret;
  }
  long long ret = 1;
  for (int i = (0); i < ((int)al[v].size()); ++i) {
    int u = al[v][i];
    if (u == p) continue;
    ret *= (dfs(u, v, 1) + dfs(u, v, 0));
    ret %= Mod;
  }
  if (!state) return dp[v][state] = ret;
  long long res = 0;
  for (int i = (0); i < ((int)al[v].size()); ++i) {
    int u = al[v][i];
    if (u == p) continue;
    long long div = dfs(u, v, 1) + dfs(u, v, 0);
    div %= Mod;
    res += (((ret * pw(div, Mod - 2)) % Mod) * dfs(u, v, 1)) % Mod;
    res %= Mod;
  }
  return dp[v][state] = res;
}
int main() {
  int n;
  cin >> n;
  for (int i = (0); i < ((int)n - 1); ++i) {
    int now;
    cin >> now;
    al[now].push_back(i + 1);
    al[i + 1].push_back(now);
  }
  for (int i = (0); i < ((int)n); ++i) {
    cin >> bad[i];
  }
  memset(dp, -1, sizeof dp);
  cout << dfs(0) << endl;
  return 0;
}
