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 + 7;
vector<int> v[maxn];
int fa[maxn], dis[maxn];
int bfs(int u) {
  memset(dis, 0x3f3f3f3f, sizeof(dis));
  queue<int> q;
  q.push(u);
  dis[u] = 0;
  int ans = 0x3f3f3f3f;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    int ln = (int)v[x].size();
    for (int i = 0; i < ln; i++) {
      int to = v[x][i];
      if (to == fa[x]) continue;
      if (dis[to] == 0x3f3f3f3f) {
        dis[to] = dis[x] + 1;
        fa[to] = x;
        q.push(to);
      } else
        ans = min(ans, dis[x] + dis[to] + 1);
    }
  }
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  bool flag = false;
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    if (flag) continue;
    for (int j = 2; j <= (int)sqrt(tmp); j++) {
      int gg = j * j;
      if (tmp % gg == 0) tmp /= gg, j--;
    }
    if (tmp == 1) {
      flag = true;
      continue;
    }
    bool f = false;
    for (int j = 2; j <= (int)sqrt(tmp); j++) {
      if (tmp % j == 0) {
        tmp /= j;
        v[j].push_back(tmp);
        v[tmp].push_back(j);
        f = true;
        break;
      }
    }
    if (!f) {
      v[1].push_back(tmp);
      v[tmp].push_back(1);
    }
  }
  int ans = 0x3f3f3f3f;
  if (flag)
    printf("1\n");
  else {
    for (int i = 1; i <= 1000; i++) {
      ans = min(ans, bfs(i));
    }
    printf("%d\n", ans == 0x3f3f3f3f ? -1 : ans);
  }
  return 0;
}