A. Two distinct points

　　A. Two distinct points

You are given two segments l1;r1 and l2;r2 on the xx-axis. It is guaranteed that l1<r1l1<r1 and l2<r2l2<r2. Segments may intersect, overlap or even coincide with each other.

The example of two segments on the xx-axis.

Your problem is to find two integers aa and bb such that l1≤a≤r1l1≤a≤r1, l2≤b≤r2l2≤b≤r2 and a≠ba≠b. In other words, you have to choose two distinct integer points in such a way that the first point belongs to the segment l1;r1 and the second one belongs to the segment l2;r2.

It is guaranteed that the answer exists. If there are multiple answers, you can print any of them.

You have to answer qq independent queries.

Input

The first line of the input contains one integer qq (1≤q≤5001≤q≤500) — the number of queries.

Each of the next qq lines contains four integers l1i,r1i,l2il1i,r1i,l2i and r2ir2i (1≤l1i,r1i,l2i,r2i≤109,l1i<r1i,l2i<r2i1≤l1i,r1i,l2i,r2i≤109,l1i<r1i,l2i<r2i) — the ends of the segments in the ii-th query.

Output

Print 2q2q integers. For the ii-th query print two integers aiai and bibi — such numbers that l1i≤ai≤r1il1i≤ai≤r1i, l2i≤bi≤r2il2i≤bi≤r2i and ai≠biai≠bi. Queries are numbered in order of the input.

It is guaranteed that the answer exists. If there are multiple answers, you can print any.

Example

input

5
1 2 1 2
2 6 3 4
2 4 1 3
1 2 1 3
1 4 5 8

output

2 1
3 4
3 2
1 2
3 7

题目描述：

在2个区间内，分别输出1个在区间里的数，2个数要不同。

分析：

从最边往里选，依次判断是否相同即可。（水题，我还打了这么多）

代码：

#include <iostream>
#include <cstdio>
using namespace std;
 
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        int l,r,L,R;
        scanf("%d%d%d%d",&l,&r,&L,&R);
        if(l==r)
        {
            printf("%d ",l);
            if(R!=r)
            printf("%d\n",R);
            else
            {
                printf("%d\n",R-1);
            }
        }
        else if(L==R)
        {
            if(l!=L)
            printf("%d ",l);
            else
            {
                printf("%d ",l+1);
            }
            printf("%d\n",R);
        }
        else 
        {
            printf("%d ",l);
            if(l!=R)
            {
                printf("%d\n",R);
            }
            else
            {
                printf("%d\n",R-1);
            }
        }
    }
    return 0;
}