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 N = 1e5 + 5;
const int M = 1e6 + 5;
int n, a[N];
vector<int> d[N], point;
vector<int> g[M];
int ans = 2e9;
void bfs(int S) {
  static int dis[M], vis[M], fa[M];
  static queue<int> q;
  dis[S] = 0;
  vis[S] = S;
  fa[S] = 0;
  q.push(S);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = 0; i < g[x].size(); ++i) {
      int y = g[x][i];
      if (vis[y] != S) {
        dis[y] = dis[x] + 1;
        vis[y] = S;
        fa[y] = x;
        q.push(y);
      } else if (fa[x] != y) {
        ans = min(ans, dis[x] + dis[y] + 1);
      }
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1, x; i <= n; ++i) {
    cin >> x;
    if (x == (int)sqrt(x) * (int)sqrt(x)) {
      cout << 1 << endl;
      return 0;
    }
    a[i] = 1;
    for (int j = 2; j * j <= x; ++j)
      if (x % j == 0) {
        int c = 0;
        while (x % j == 0) x /= j, c++;
        if (c & 1) {
          d[i].push_back(j);
          a[i] *= j;
        }
      }
    if (x > 1) {
      d[i].push_back(x);
      a[i] *= x;
    }
    if (d[i].size() == 1) d[i].push_back(1);
  }
  for (int i = 1; i <= n; ++i) {
    g[d[i][0]].push_back(d[i][1]);
    g[d[i][1]].push_back(d[i][0]);
    point.push_back(d[i][0]);
    point.push_back(d[i][1]);
  }
  sort(point.begin(), point.end());
  point.erase(unique(point.begin(), point.end()), point.end());
  for (int i = 0; i < point.size(); ++i)
    if (point[i] <= 1e3) {
      bfs(point[i]);
    }
  if (ans <= n)
    cout << ans << endl;
  else
    cout << -1 << endl;
}