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 MAX_N = 101000;
const long long MOD = 1e9 + 7;
vector<int> v[MAX_N];
int col[MAX_N];
long long dp[MAX_N][2];
bool have[MAX_N];
long long ans = 1;
void dfs(int x, int fa) {
  int i;
  dp[x][0] = 1;
  dp[x][1] = 0;
  for (i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (y == fa) continue;
    dfs(y, x);
    dp[x][1] = (dp[x][1] * dp[y][0] + dp[x][0] * dp[y][1]) % MOD;
    dp[x][0] = dp[x][0] * dp[y][0] % MOD;
  }
  if (col[x] == 1)
    dp[x][1] = dp[x][0];
  else
    dp[x][0] = (dp[x][1] + dp[x][0]) % MOD;
}
int main(void) {
  int n, x, i;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d", &x);
    x++;
    v[x].push_back(i + 1);
    v[i + 1].push_back(x);
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &col[i]);
  }
  dfs(1, 0);
  cout << dp[1][1] << "\n";
  return 0;
}
