You're given an array a of length 2n. Is it possible to reorder it in such way so that the sum of the first n elements isn't equal to the sum of the last n

elements?

Input

The first line contains an integer n

(1≤n≤1000), where 2n is the number of elements in the array a

.

The second line contains 2n

space-separated integers a1, a2, …, a2n (1≤ai≤106) — the elements of the array a

.

Output

If there's no solution, print "-1" (without quotes). Otherwise, print a single line containing 2n

space-separated integers. They must form a reordering of a

. You are allowed to not change the order.

Examples

Input

Copy

3
1 2 2 1 3 1

Output

Copy

2 1 3 1 1 2

Input

Copy

1
1 1

Output

Copy

-1

Note

In the first example, the first n

elements have sum 2+1+3=6 while the last n elements have sum 1+1+2=4

. The sums aren't equal.

In the second example, there's no solution.

思路：排个序，前一半的和如果和后一半不同，输出排序后的数组，相同输出-1

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define Max 105
#include<queue>
int main()
{
    int n;
    scanf("%d",&n);

        int a[2200]={0};
        for(int i=1;i<=2*n;i++) scanf("%d",&a[i]);
        sort(a+1,a+2*n+1);
        int x=0,y=0;
        for(int i=1;i<=n;i++) x+=a[i];
        for(int i=n+1;i<=2*n;i++)
        {
            y+=a[i];
        }
        if(x==y)
        {
            printf("-1\n");

        }
        else
        {
            printf("%d",a[1]);
            for(int i=2;i<=2*n;i++) printf(" %d",a[i]);
            printf("\n");
        }

}