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 MX = 2 * 1e5 + 11;
int N, M, k;
char s[MX];
long long dp[MX][2];
vector<int> G[MX];
int arr[MX];
long long md = 1e9 + 7;
void dfs(int x, int p) {
  dp[x][0] = arr[x] == 0;
  dp[x][1] = arr[x] == 1;
  for (int i = 0; i < G[x].size(); i++) {
    int ch = G[x][i];
    if (ch == p) continue;
    dfs(ch, x);
    dp[x][1] = dp[x][1] * (dp[ch][1] + dp[ch][0]);
    dp[x][1] %= md;
    dp[x][1] += dp[x][0] * dp[ch][1];
    dp[x][1] %= md;
    dp[x][0] = dp[x][0] * (dp[ch][0] + dp[ch][1]);
    dp[x][0] %= md;
  }
}
int main() {
  cin >> N;
  for (int i = 2; i <= N; i++) {
    int a;
    scanf("%d", &a);
    G[i].push_back(++a);
    G[a].push_back(i);
  }
  for (int i = 1; i <= N; i++) scanf("%d", &arr[i]);
  dfs(1, -1);
  cout << dp[1][1] << "\n";
}
