Codeforces 1206D 思维+floyed求最小环

D. Shortest Cycle

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given nn integer numbers a1,a2,…,ana1,a2,…,an. Consider graph on nn nodes, in which nodes ii, jj (i≠ji≠j) are connected if and only if, aiaiAND aj≠0aj≠0, where AND denotes the 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 nn (1≤n≤105)(1≤n≤105) — number of numbers.

The second line contains nn integer numbers a1,a2,…,ana1,a2,…,an (0≤ai≤10180≤ai≤1018).

Output

If the graph doesn't have any cycles, output −1−1. 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)(9,3,6,28).

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

The graph has no cycles in the third example.

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 100010
using namespace std;
int n,map[210][210],dis[210][210],ans;
long long a[maxn];
int main(){
    scanf("%d",&n);
    int cnt=0;
    for(int i=1;i<=n;i++){
        long long x;
        scanf("%lld",&x);
        if(x!=0)a[++cnt]=x;
    }
    n=cnt;
    if(n>=200){
        puts("3");
        return 0;
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            if(i==j)continue;
            dis[i][j]=map[i][j]=100000000;
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            if(i==j)continue;
            if((a[i]&a[j])!=0)
                map[i][j]=map[j][i]=dis[i][j]=dis[j][i]=1;
        }
    ans=0x7fffffff;
    for(int k=1;k<=n;k++){
        for(int i=1;i<k;i++){
            for(int j=i+1;j<k;j++){
                ans=min(dis[i][j]+map[i][k]+map[k][j],ans);
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
            }
        }
    }
    if(ans>n){
        puts("-1");
        return 0;
    }
    else printf("%d\n",ans);
    return 0;
}