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 n, clr[100005];
vector<int> adj[100005];
long long dp[100005][2];
long long bigmod(long long x, int pow) {
  long long ret = 1;
  while (pow > 0) {
    if (pow & 1) ret = (ret * x) % 1000000007;
    x = (x * x) % 1000000007;
    pow /= 2;
  }
  return ret;
}
long long dfs(int node, int par, int val) {
  if (dp[node][val] != -1) return dp[node][val];
  vector<int> v0, v1, Ts;
  if (!val && clr[node]) return 0;
  for (int i = 0; i < (int)adj[node].size(); i++) {
    int nd = adj[node][i];
    if (nd == par) continue;
    v0.push_back(dfs(nd, node, 0));
    v1.push_back(dfs(nd, node, 1));
  }
  long long ret = 0, whole = 1;
  Ts.push_back(1);
  for (int i = 0; i < (int)v0.size(); i++) {
    whole = (v0[i] + v1[i]) * whole;
    whole %= 1000000007;
    Ts.push_back(whole);
  }
  if (val && !clr[node]) {
    long long tmp = 1;
    for (int i = (int)v0.size() - 1; i >= 0; i--) {
      long long a = (Ts[i] * tmp) % 1000000007;
      a = (a * v1[i]) % 1000000007;
      ret = (ret + a) % 1000000007;
      tmp = tmp * (v0[i] + v1[i]);
      tmp %= 1000000007;
    }
  } else
    ret = whole;
  return dp[node][val] = ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < (int)n - 1; i++) {
    int u;
    cin >> u;
    adj[u].push_back(i + 1);
    adj[i + 1].push_back(u);
  }
  for (int i = 0; i < (int)n; i++) cin >> clr[i];
  memset(dp, -1, sizeof(dp));
  cout << dfs(0, -1, 1) << endl;
}
