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 long long N = 100100;
const long long Mod = 1e9 + 7;
long long Rec(long long i, long long j, long long c);
long long p[N];
long long x[N];
vector<long long> Adj[N];
vector<long long> Dp[N][2];
int main() {
  long long n;
  cin >> n;
  for (long long i = 2; i <= n; i++)
    scanf("%I64d", p + i), Adj[++p[i]].push_back(i);
  for (long long i = 1; i <= n; i++) scanf("%I64d", x + i);
  for (long long i = 0; i < N; i++)
    Dp[i][0] = Dp[i][1] = vector<long long>(Adj[i].size(), -1);
  cout << Rec(1, 0, 0);
}
long long Rec(long long i, long long j, long long c) {
  if (j == 0) c += x[i];
  if (c > 1) return 0;
  if (j == Adj[i].size()) return c;
  if (Dp[i][c][j] + 1) return Dp[i][c][j];
  long long Res1 = (Rec(i, j + 1, c) * Rec(Adj[i][j], 0, 0)) % Mod;
  long long Res2 = (c ? Rec(i, j + 1, 1) * Rec(Adj[i][j], 0, 1) : 0) % Mod;
  long long Res3 = (!c ? (Rec(i, j + 1, 1) * Rec(Adj[i][j], 0, 0)) % Mod +
                             (Rec(i, j + 1, 0) * Rec(Adj[i][j], 0, 1)) % Mod
                       : 0) %
                   Mod;
  return Dp[i][c][j] = (Res1 + Res2 + Res3) % Mod;
}
