## Problem – Best of Two CodeChef Solution

Chef took an examination two times. In the first attempt, he scored X*X* marks while in the second attempt he scored Y*Y* marks. According to the rules of the examination, the best score out of the two attempts will be considered as the final score.

Determine the final score of the Chef.

#### Input Format

- The first line contains a single integer T
*T* — the number of test cases. Then the test cases follow. - The first line of each test case contains two integers X
*X* and Y*Y* — the marks scored by Chef in the first attempt and second attempt respectively.

#### Output Format

For each test case, output the final score of Chef in the examination.

#### Constraints

- 1 \leq T \leq 10001≤
*T*≤1000 - 0 \le X, Y \le 1000≤
*X*,*Y*≤100

#### Sample 1:

```
Input: 4
40 60
67 55
50 50
1 100
Output: 60
67
50
100
```

### Explanation:

**Test Case 1:** The best score out of the two attempts is 6060.

**Test Case 2:** The best score out of the two attempts is 6767.

**Test Case 3:** The best score out of the two attempts is 5050.

### Best of Two CodeChef Solution in Python3

```
for i in range(int(input())):
a,b=map(int,input().split())
if b>a:
print(b)
else:
print(a)
```

### Best of Two CodeChef Solution in C++14

```
#include <iostream>
using namespace std;
int main() {
int n;
cin>>n;
while(n--){
int x,y;
cin>>x>>y;
if (x>y)
cout<<x<<endl;
else
cout<<y<<endl;
}
return 0;
}
```

### Best of Two CodeChef Solution in Java

```
import java.util.*;
import java.lang.*;
import java.io.*;
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i=0;i<t ;i++ ){
int x = sc.nextInt();
int y = sc.nextInt();
if(x>=y){
System.out.println(x);
}
else{
System.out.println(y);
}
}
}
}
```

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