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 long long int inf = 1e18, M = 1e9 + 7;
const long long int N = 1e5 + 5;
vector<long long int> v[N], val(N, 0);
long long int dp[N][2];
long long int zero, one;
void dfs(long long int x, long long int p) {
  dp[x][val[x]] = 1;
  for (auto c : v[x]) {
    if (c == p) continue;
    dfs(c, x);
    zero = dp[x][0];
    one = dp[x][1];
    dp[x][0] = (zero * dp[c][0]) % M;
    dp[x][1] = ((zero * dp[c][1]) % M + (one * dp[c][0]) % M) % M;
    dp[x][0] = (dp[x][0] + (zero * dp[c][1]) % M) % M;
    dp[x][1] = (dp[x][1] + (one * dp[c][1]) % M) % M;
  }
}
void solve() {
  memset(dp, 0, sizeof(dp));
  long long int n;
  cin >> n;
  long long int x;
  for (long long int i = 1; i < n; ++i) {
    cin >> x;
    v[x].push_back(i);
    v[i].push_back(x);
  }
  for (long long int i = 0; i < n; ++i) cin >> val[i];
  dfs(0, -1);
  cout << dp[0][1];
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
