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 N = 1e5 + 7;
const int MOD = 1e9 + 7;
int dp[N][2];
int col[N];
vector<int> child[N];
long long bpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
void dfs(int v) {
  long long mul = 1;
  for (auto u : child[v]) {
    dfs(u);
    mul = mul * (dp[u][1] + dp[u][0]) % MOD;
  }
  if (col[v])
    dp[v][1] = mul;
  else {
    dp[v][0] = mul;
    dp[v][1] = 0;
    for (auto u : child[v])
      dp[v][1] = (dp[v][1] + mul * bpow(dp[u][1] + dp[u][0], MOD - 2) % MOD *
                                 dp[u][1] % MOD) %
                 MOD;
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int p;
    cin >> p;
    child[p].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
  dfs(0);
  cout << dp[0][1] << "\n";
  return 0;
}
