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 factor[1010000];
int a[1010000];
vector<int> g[1010000];
int n;
int depth[1010000];
int ans = 0x3f3f3f3f;
int visit[1010000];
int lst[1010000];
int main() {
  factor[1] = 1;
  for (int i = 2; i <= 1000000; i++)
    if (factor[i] == 0)
      for (int j = i; j <= 1000000; j += i)
        if (!factor[j]) factor[j] = i;
  cin >> n;
  int x, y;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    x = factor[a[i]];
    cnt = 0;
    while (factor[a[i]] == x && x != 1) {
      a[i] /= x;
      cnt++;
    }
    x = (cnt & 1) ? x : 1;
    cnt = 0;
    y = factor[a[i]];
    while (factor[a[i]] == y && y != 1) {
      a[i] /= y;
      cnt++;
    }
    y = (cnt & 1) ? y : 1;
    if (x == y) {
      cout << 1;
      return 0;
    }
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= 1000; i++) {
    visit[i] = i;
    depth[i] = 0;
    lst[i] = 0;
    queue<int> q;
    q.push(i);
    while (!q.empty()) {
      int pos = q.front();
      q.pop();
      for (auto p : g[pos])
        if (p != lst[pos])
          if (visit[p] == i)
            ans = min(ans, depth[pos] + depth[p] + 1);
          else {
            visit[p] = i;
            depth[p] = depth[pos] + 1;
            lst[p] = pos;
            q.push(p);
          }
    }
  }
  if (ans == 0x3f3f3f3f)
    cout << -1;
  else
    cout << ans;
  return ~~(0 ^ 0 ^ 0);
}