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 int MOD = 1e9 + 7;
int n, i, j, mark[100005];
long long dp[2][100005];
vector<int> edge[100005];
void solve(int id, int p) {
  for (int k = 0; k < edge[id].size(); k++) {
    int to = edge[id][k];
    if (to == p) continue;
    solve(to, id);
  }
  if (mark[id]) {
    long long pd = 1LL;
    for (int k = 0; k < edge[id].size(); k++) {
      int to = edge[id][k];
      if (to == p) continue;
      pd = (pd * dp[0][to]) % MOD;
    }
    dp[1][id] = pd;
    dp[0][id] = 0;
  } else {
    long long pd = 1LL;
    vector<long long> num;
    vector<long long> l;
    vector<long long> r;
    for (int k = 0; k < edge[id].size(); k++) {
      int to = edge[id][k];
      if (to == p) continue;
      pd = (pd * dp[0][to]) % MOD;
      num.push_back(dp[1][to]);
      l.push_back(dp[0][to]);
      r.push_back(dp[0][to]);
    }
    dp[0][id] = pd;
    for (int k = 1; k < l.size(); k++) l[k] = (l[k] * l[k - 1]) % MOD;
    for (int k = r.size() - 2; k >= 0; k--) r[k] = (r[k] * r[k + 1]) % MOD;
    for (int k = 0; k < l.size(); k++) {
      long long now = num[k];
      if (k - 1 >= 0) now = (now * l[k - 1]) % MOD;
      if (k + 1 < r.size()) now = (now * r[k + 1]) % MOD;
      dp[1][id] = (dp[1][id] + now) % MOD;
    }
  }
  if (p != -1) dp[0][id] = (dp[0][id] + dp[1][id]) % MOD;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    int x;
    scanf("%d", &x);
    edge[x].push_back(i);
    edge[i].push_back(x);
  }
  for (i = 0; i < n; i++) scanf("%d", &mark[i]);
  solve(0, -1);
  printf("%lld\n", dp[1][0]);
}
