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;
int n, kq = 0;
vector<int> a[100100];
int x[100100], dd[100100];
long long g[100100], first[100100];
int d[100100][3];
void nhap() {
  cin >> n;
  int c;
  for (int i = 0; i <= n - 2; i++) {
    scanf("%d", &c);
    a[c].push_back(i + 1);
    a[i + 1].push_back(c);
  }
  for (int i = 0; i <= n - 1; i++) scanf("%d", &x[i]);
}
void chuanbi() {
  memset(g, 0, sizeof(g));
  memset(first, 0, sizeof(first));
  memset(dd, 0, sizeof(dd));
}
void dfs(int v) {
  dd[v] = 1;
  if (!x[v])
    first[v] = 1;
  else
    g[v] = 1;
  for (int i = 0; i < a[v].size(); i++) {
    int u = a[v][i];
    if (!dd[u]) {
      dfs(u);
      g[v] = (g[v] * (first[u] + g[u]) % 1000000007 +
              first[v] * g[u] % 1000000007) %
             1000000007;
      first[v] = first[v] * (first[u] + g[u]) % 1000000007;
    }
  }
}
void xuli() {
  dfs(1);
  cout << g[1];
}
void ghikq() {}
int main() {
  nhap();
  chuanbi();
  xuli();
  ghikq();
}
