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 mxmx = 1e5 + 10;
const long long int mod = 1e9 + 7;
long long int n;
vector<long long int> adj[mxmx];
bool color[mxmx];
long long int dp[2][mxmx];
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int ANS;
  ANS = power(a, b / 2);
  ANS *= ANS;
  ANS %= mod;
  ANS *= power(a, b % 2);
  ANS %= mod;
  return ANS;
}
void go(long long int v) {
  if (color[v] == 0) {
    dp[1][v] = 0;
    dp[0][v] = 1;
  } else {
    dp[0][v] = 0;
    dp[1][v] = 1;
  }
  long long int next;
  long long int zarb = 1;
  for (long long int i = 0; i < adj[v].size(); i++) {
    next = adj[v][i];
    go(next);
    zarb *= ((dp[0][next] + dp[1][next]) % mod);
    zarb %= mod;
  }
  long long int A, B;
  for (long long int i = 0; i < adj[v].size(); i++) {
    next = adj[v][i];
    if (color[v] == 0) {
      dp[0][v] *= ((dp[1][next] + dp[0][next]) % mod);
      dp[0][v] %= mod;
      A = (zarb * dp[1][next]) % mod;
      B = (dp[0][next] + dp[1][next]) % mod;
      dp[1][v] += (A * power(B, mod - 2)) % mod;
      dp[1][v] %= mod;
    } else {
      dp[0][v] = 0;
      dp[1][v] *= ((dp[1][next] + dp[0][next]) % mod);
      dp[1][v] %= mod;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long int i = 1, p; i < n; i++) {
    cin >> p;
    adj[p].push_back(i);
  }
  for (long long int i = 0; i < n; i++) {
    cin >> color[i];
  }
  go(0);
  cout << dp[1][0];
}
