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 + 6;
int h[MAXN], par[MAXN], a[MAXN];
set<int> adj[MAXN];
queue<int> q;
int n, ans = INT_MAX;
void input();
void bfs(int);
void solve();
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input();
  solve();
  return 0;
}
void input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    int u, v;
    u = v = 1;
    for (int j = 2; j * j <= a[i]; j++) {
      int x = j, cnt = 0;
      while (a[i] % x == 0) x *= j, cnt++;
      x /= j;
      a[i] /= x;
      if (cnt % 2) u = v, v = j;
    }
    if (a[i] > 1) u = v, v = a[i];
    if (u == 1 && v == 1) ans = 1;
    if (adj[u].count(v)) ans = min(ans, 2);
    adj[u].insert(v);
    adj[v].insert(u);
  }
}
void bfs(int u) {
  memset(h, 63, sizeof h);
  h[u] = 0;
  q.push(u);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (auto v : adj[x])
      if (h[v] > 1e6) {
        h[v] = h[x] + 1;
        par[v] = x;
        q.push(v);
      } else if (v != par[x])
        ans = min(ans, h[x] + h[v] + 1);
  }
}
void solve() {
  for (int i = 1; i < 1000; i++)
    if (adj[i].size()) bfs(i);
  if (ans == INT_MAX) {
    cout << -1;
    return;
  }
  cout << ans;
}