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>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double PI = acos(-1);
const long long MOD = 1000000007;
const long long FMOD = 998244353;
const long double eps = 1e-9;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
vector<long long> v[100005];
long long c[100005];
long long dp[2][100005];
void dfs(long long src, long long par) {
  long long zero = 1 - c[src];
  long long one = c[src];
  for (auto x : v[src]) {
    if (x != par) {
      dfs(x, src);
      one = (one * dp[0][x] + one * dp[1][x] + zero * dp[1][x]) % MOD;
      zero = (zero * dp[0][x] + zero * dp[1][x]) % MOD;
    }
  }
  dp[0][src] = zero;
  dp[1][src] = one;
}
signed main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long a;
    cin >> a;
    v[a].push_back(i + 1);
    v[i + 1].push_back(a);
  }
  for (long long i = 0; i < n; i++) cin >> c[i];
  dfs(0, -1);
  cout << dp[1][0];
}
