Codeforces Round #580 (Div. 2)-D. Shortest Cycle（思维建图+dfs找最小环）

You are given bitwise AND operation.

Find the length of the shortest cycle in this graph or determine that it doesn't have cycles at all.

Input

The first line contains one integer (1≤n≤105) — number of numbers.

The second line contains 0≤ai≤1018).

Output

If the graph doesn't have any cycles, output −1. Else output the length of the shortest cycle.

Examples

input

Copy

4
3 6 28 9

output

Copy

4

input

Copy

5
5 12 9 16 48

output

Copy

3

input

Copy

4
1 2 4 8

output

Copy

-1

Note

In the first example, the shortest cycle is (9,3,6,28).

In the second example, the shortest cycle is (5,12,9).

The graph has no cycles in the third example.

 代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>
#define lson m<<1,l,mid
#define rson m<<1|1,mid+1,r
#define getmid(m) (tree[m].l+tree[m].r)>>1;
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
pair<int,int>pp;
vector<int>vec[65];
vector<int>G[maxn];
ll a[maxn];
int vis[maxn];
int endd;
int minn=0x3f3f3f3f;
void  dfs(int sta,int dep)
{
    vis[sta]=1;
    if(dep>=minn)
    {
          vis[sta]=0;
          return;
    }
    for(int t=0;t<G[sta].size();t++)
    {  
      int u=G[sta][t];
      if(u==endd&&dep>1)
      {
          minn=min(minn,dep+1);
      }
      if(vis[u]==0)
      {
          dfs(u,dep+1);
      }
    }
    vis[sta]=0;  
}
 
int main()
{
    int n;
    cin>>n;
    for(int t=0;t<n;t++)
    {
        scanf("%lld",&a[t]);
    }
    for(int t=0;t<64;t++)
    {
        for(int j=0;j<n;j++)
        {
            if(a[j]&((1ll<<t)))
            {
             vec[t].push_back(j);
            }
        }
    }
    for(int t=0;t<64;t++)
    {
        if(vec[t].size()>=3)
        {
            puts("3");
            return 0;
        }
    }
    for(int t=0;t<64;t++)
    {
        if(vec[t].size()==2)
        {
            int u=vec[t][0];
            int v=vec[t][1];
            G[u].push_back(v);
            G[v].push_back(u);
        }
    }
    for(int t=0;t<n;t++)
    {
        endd=t;
        dfs(t,0);
    }
    if(minn==0x3f3f3f3f)
    {
        puts("-1");
    }
    else
    {
    printf("%d\n",minn);
    }
   // system("pause");
    return 0;
}