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 + 10;
struct node {
  int u;
  int v;
  int next;
  node(int U = 0, int V = 0, int N = 0) : u(U), v(V), next(N) {}
} e[5000001];
int first[5000011], cnt, ans = 0x3f3f3f3f, inf = 0x3f3f3f3f;
void add(int u, int v) {
  e[++cnt] = node(u, v, first[u]);
  first[u] = cnt;
  e[++cnt] = node(v, u, first[v]);
  first[v] = cnt;
}
int book[maxn], pre[maxn];
void bfs(int u) {
  memset(book, inf, sizeof(book));
  book[u] = 0;
  queue<int> q;
  q.push(u);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = first[u]; i; i = e[i].next) {
      int v = e[i].v;
      if (v != pre[u]) {
        if (book[v] != inf)
          ans = min(ans, (book[u] + book[v] + 1));
        else
          pre[v] = u, book[v] = book[u] + 1, q.push(v);
        ;
      }
    }
  }
}
int main() {
  int n, x;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x;
    for (int i = 2; i <= sqrt(x); i++) {
      while (x % (i * i) == 0) x /= i * i;
    }
    if (x == 1) {
      cout << 1 << endl;
      return 0;
    }
    int u = x, v = 1;
    add(u, v);
    for (int i = 2; i <= sqrt(x); i++) {
      if (x % i == 0) {
        u = x / i, v = i;
        add(u, v);
        break;
      }
    }
  }
  for (int i = 1; i <= sqrt(maxn); i++) bfs(i);
  if (ans != inf)
    cout << ans << endl;
  else
    printf("-1");
  return 0;
}