Problem: You are given an array a of length n that has a special condition: every element in this array has at most 7 divisors. Find the length of the shortest non-empty subsequence of this array product of whose elements is a perfect square.

A sequence a is a subsequence of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5) — the length of a.

The second line contains n integers a_1, a_2, …, a_{n} (1 ≤ a_i ≤ 10^6) — the elements of the array a.

Output

Output the length of the shortest non-empty subsequence of a product of whose elements is a perfect square. If there are several shortest subsequences, you can find any of them. If there's no such subsequence, print "-1".

Examples

Input


3
1 4 6


Output


1

Input


4
2 3 6 6


Output


2

Input


3
6 15 10


Output


3

Input


4
2 3 5 7


Output


-1

Note

In the first sample, you can choose a subsequence [1].

In the second sample, you can choose a subsequence [6, 6].

In the third sample, you can choose a subsequence [6, 15, 10].

In the fourth sample, there is no such subsequence.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, x, ans;
pair<int, int> v[100005];
vector<int> g[1000005];
int f[1000005], t[1000005];
void dfs(int nod) {
  f[nod] = (int)1e9;
  t[nod] = -1;
  for (auto &fiu : g[nod]) {
    if (f[fiu] != (int)1e9) dfs(fiu);
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  ans = n + 5;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    v[i].first = 1;
    for (int j = 2; j * j <= x; j++) {
      int e = 0;
      while (x % j == 0) e++, x /= j;
      if (e % 2) v[i] = {v[i].first * j, j};
    }
    if (x > 1) v[i] = {v[i].first * x, x};
    if (v[i].first == 1) {
      cout << 1;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (f[v[i].first]) {
      cout << 2;
      return 0;
    }
    f[v[i].first] = 1;
  }
  for (int i = 1; i <= n; i++) {
    g[v[i].first / v[i].second].push_back(v[i].second);
    g[v[i].second].push_back(v[i].first / v[i].second);
  }
  for (int i = 1; i <= 1000000; i++) f[i] = 0;
  for (int i = 1; i <= 1000; i++) {
    queue<int> q;
    dfs(i);
    q.push(i);
    f[i] = 0;
    while (!q.empty()) {
      int nod = q.front();
      q.pop();
      for (auto &fiu : g[nod]) {
        if (f[fiu] > f[nod] + 1) {
          f[fiu] = f[nod] + 1;
          t[fiu] = nod;
          q.push(fiu);
        } else if (t[fiu] != nod && t[nod] != fiu)
          ans = min(ans, f[fiu] + f[nod] + 1);
      }
    }
  }
  if (ans != n + 5)
    cout << ans;
  else
    cout << -1;
  return 0;
}