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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* na, Arg1&& arg1, Args&&... args) {
  const char* c = strchr(na + 1, ',');
  cerr.write(na, c - na) << " : " << arg1 << " , ";
  __f(c + 1, args...);
}
const int MOD = 1e9 + 7;
const int inf = 1e9;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const long long INF = 1e18;
const int N = 1e5 + 5;
int col[N];
vector<int> graph[N];
long long int dp[N][2];
void dfs(int s, int p) {
  dp[s][0] = 1;
  dp[s][1] = 0;
  for (auto x : graph[s]) {
    if (x == p) continue;
    dfs(x, s);
    dp[s][1] = (dp[s][1] % MOD * dp[x][0] % MOD) % MOD;
    dp[s][1] = (dp[s][1] % MOD + (dp[s][0] % MOD * dp[x][1] % MOD) % MOD) % MOD;
    dp[s][0] = (dp[s][0] % MOD * dp[x][0] % MOD) % MOD;
  }
  if (col[s] == 1)
    dp[s][1] = dp[s][0] % MOD;
  else
    dp[s][0] = (dp[s][0] % MOD + dp[s][1] % MOD) % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u;
    cin >> u;
    graph[u].push_back(i);
    graph[i].push_back(u);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
  dfs(0, -1);
  cout << dp[0][1] % MOD << "\n";
  return 0;
}
