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;
const int MAXN = 1e6 + 7;
const int inf = 0x3f3f3f3f;
int ans = inf;
int par[MAXN];
int d[MAXN];
vector<int> g[MAXN];
void bfs(int p) {
  memset(d, inf, sizeof d);
  queue<int> q;
  q.push(p);
  d[p] = 0;
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (auto v : g[u]) {
      if (v != par[u]) {
        if (d[v] == inf) {
          par[v] = u;
          d[v] = d[u] + 1;
          q.push(v);
        } else {
          ans = min(ans, d[v] + d[u] + 1);
        }
      }
    }
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    vector<int> primes;
    for (int j = 2; j * j <= a[i]; j++) {
      int cnt = 0;
      while (a[i] % j == 0) a[i] /= j, cnt++;
      if (cnt % 2 == 1) primes.push_back(j);
    }
    if (a[i] != 1) {
      primes.push_back(a[i]);
    }
    if (primes.empty()) {
      cout << 1 << endl;
      return;
    }
    if (primes.size() == 1) {
      primes.push_back(1);
    }
    assert(primes.size() == 2);
    g[primes[0]].push_back(primes[1]);
    g[primes[1]].push_back(primes[0]);
  }
  for (int i = 1; i <= 1000; i++) bfs(i);
  cout << (ans == inf ? -1 : ans) << endl;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}