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 mod;
int n;
map<int, vector<int> > ch;
int col[100001];
bool seen[100001];
int dp[100001][2];
void dfs(int r) {
  if (seen[r]) return;
  seen[r] = true;
  dp[r][0] = 1;
  dp[r][1] = 0;
  for (int i = 0; i < ch[r].size(); ++i) {
    int u = ch[r][i];
    if (seen[u]) continue;
    dfs(u);
    dp[r][1] = ((dp[r][1] * 1ll) * dp[ch[r][i]][0]) % mod;
    dp[r][1] =
        (dp[r][1] * 1ll + ((1ll * dp[ch[r][i]][1]) * dp[r][0]) % mod) % mod;
    dp[r][0] = ((dp[r][0] * 1ll) * dp[ch[r][i]][0]) % mod;
  }
  if (col[r] == 1) {
    dp[r][1] = dp[r][0];
  } else {
    dp[r][0] = (dp[r][0] * 1ll + dp[r][1]) % mod;
  }
}
int main() {
  mod = 1000000007;
  memset(dp, 0, sizeof dp);
  memset(seen, false, sizeof seen);
  scanf("%d\n", &n);
  int tmp;
  for (int i = 0; i < n - 1; ++i) {
    cin >> tmp;
    ch[tmp].push_back(i + 1);
    ch[i + 1].push_back(tmp);
  }
  cin.get();
  for (int i = 0; i < n; ++i) {
    cin >> col[i];
  }
  dfs(0);
  cout << dp[0][1] << endl;
  return 0;
}
