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 mod = 1000000007, M = 1e5 + 7;
long long powe(long long x, long long y) {
  x = x % mod;
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
vector<int> edg[M];
bool col[M], vis[M];
long long s0[M], s1[M];
void dfs(int i) {
  vis[i] = true;
  if (col[i])
    s1[i] = 1;
  else
    s0[i] = 1;
  for (auto x : edg[i]) {
    if (vis[x]) continue;
    dfs(x);
    s1[i] = (((s1[i] * s0[x])) % mod + (s0[i] * s1[x]) % mod +
             (s1[i] * s1[x]) % mod) %
            mod;
    s0[i] = ((s0[i] * s0[x]) % mod + (s0[i] * s1[x]) % mod) % mod;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, a;
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> a;
    edg[i].push_back(a);
    edg[a].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
  dfs(0);
  cout << s1[0];
  return 0;
}
