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>
std::vector<int> children[100005];
int N;
long long int dp1[100005], dp2[100005];
bool isBlack[100005];
inline int max(int a, int b) { return a + (b - a) * (b > a); }
void extendedGCD(int a, int b, long long int& x, long long int& y) {
  if (b) {
    extendedGCD(b, a % b, x, y);
    int aux = y;
    y = x - y * (a / b);
    x = aux;
  } else {
    x = 1;
    y = 0;
  }
}
int modullarInverse(int x) {
  long long int a, b;
  extendedGCD(1000000007, x, a, b);
  while (b < 0) b += 1000000007;
  b %= 1000000007;
  return b;
}
void solve(int node) {
  long long int prod = 1;
  for (std::vector<int>::iterator i0 = children[node].begin(),
                                  i1 = children[node].end();
       i0 != i1; ++i0) {
    solve(*i0);
    if (dp2[*i0]) prod = (prod * (dp1[*i0] + dp2[*i0])) % 1000000007;
  }
  if (!isBlack[node]) {
    dp1[node] = 1;
    dp2[node] = 0;
    for (std::vector<int>::iterator i0 = children[node].begin(),
                                    i1 = children[node].end();
         i0 != i1; ++i0) {
      dp1[node] = (dp1[node] * (dp1[*i0] + dp2[*i0])) % 1000000007;
      if (dp2[*i0])
        dp2[node] = (dp2[node] + (prod * modullarInverse(dp1[*i0] + dp2[*i0]) %
                                  1000000007) *
                                     dp2[*i0]) %
                    1000000007;
    }
  } else
    dp2[node] = prod, dp1[node] = 0;
}
int main() {
  int i, x;
  scanf("%d", &N);
  for (i = 1; i < N; ++i) {
    scanf("%d", &x);
    children[x].push_back(i);
  }
  for (i = 0; i < N; ++i) scanf("%d", isBlack + i);
  solve(0);
  printf("%lld", dp2[0]);
  return 0;
}
