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;
vector<int> G[100005];
long long dp[100005][2];
int c[100005];
long long ksm(long long a, long long b) {
  long long c = 1, d = a;
  while (b) {
    if (b & 1) c = (c * d) % 1000000007;
    d = (d * d) % 1000000007;
    b >>= 1;
  }
  return c;
}
void dfs(int x) {
  int i, len, v;
  len = G[x].size();
  dp[x][0] = dp[x][1] = 1;
  for (i = 0; i < len; i++) {
    v = G[x][i];
    dfs(v);
    if (c[x] == 0)
      dp[x][0] = dp[x][0] * (dp[v][0] + dp[v][1]) % 1000000007;
    else
      dp[x][1] = dp[x][1] * (dp[v][0] + dp[v][1]) % 1000000007;
  }
  if (c[x] == 1)
    dp[x][0] = 0;
  else {
    dp[x][1] = 0;
    for (i = 0; i < len; i++) {
      v = G[x][i];
      dp[x][1] = (dp[x][1] + (dp[x][0] * dp[v][1] % 1000000007) *
                                 ksm(dp[v][0] + dp[v][1], 1000000007 - 2)) %
                 1000000007;
    }
  }
}
int main() {
  int n, i, x;
  cin >> n;
  for (i = 2; i <= n; i++) {
    scanf("%d", &x);
    G[x + 1].push_back(i);
  }
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  dfs(1);
  cout << dp[1][1] % 1000000007 << endl;
  return 0;
}
