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 MAXA = 1000100, INF = 998244353;
int n, ans = INF, lvl[MAXA], p[MAXA];
vector<int> g[MAXA];
void bfs(int v) {
  for (int i = 0; i < MAXA; i++) lvl[i] = INF;
  queue<int> q;
  lvl[v] = 0;
  q.push(v);
  while (!q.empty()) {
    v = q.front(), q.pop();
    for (const int &u : g[v])
      if (u != p[v]) {
        if (lvl[u] == INF)
          p[u] = v, lvl[u] = lvl[v] + 1, q.push(u);
        else
          ans = min(ans, lvl[u] + lvl[v] + 1);
      }
  }
}
void f(int x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) {
      while (x % (i * i) == 0) x /= i * i;
    }
  if (x == 1) ans = 1;
  int p1 = x, p2 = 1;
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) p1 = x / i, p2 = i;
  g[p1].push_back(p2);
  g[p2].push_back(p1);
}
int main() {
  cin >> n;
  for (int i = 1, x; i <= n; ++i) cin >> x, f(x);
  for (int i = 1; i * i < MAXA; ++i) bfs(i);
  if (ans == INF)
    cout << -1;
  else
    cout << ans << endl;
}