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Chef has 2 numbers A and B (A<B).

Chef will perform some operations on A.

In the ith operation:

• Chef will add 1 to A if i is odd.
• Chef will add 2 to A if i is even.

Chef can stop at any instant. Can Chef make A equal to B?

Input Format

• The first line contains a single integer T — the number of test cases. Then the test cases follow.
• The first and only line of each test case contains two space separated integers A and B.

Output Format

For each test case, output `YES` if Chef can make A and B equal, `NO` otherwise.

Note that the checker is case-insensitive. So, `YES``Yes``yEs` are all considered same.

• 1≤T≤1000
• 1≤A<B≤109

Input

```4
1 2
3 6
4 9
10 20
```

Output

```YES
YES
NO
YES
```

SOLUTION

Program: Alternate Additions Solution in Python

```for _ in  range(int(input())):
a,b = map(int,input().split(' '))
N = b-a
if(N%3==0 or N%3==1):
print('yes')
else:
print('No')```

Program: Alternate Additions Solution in C++

```#include <iostream>
using namespace std;

int main()
{
int t;
cin>>t;
while(t--)
{
int a,b;
cin>>a>>b;
if(b-a<0)
{
cout<<"NO"<<endl;
}
else
{
if((b-a)%3==0||(b-a-1)%3==0)
{
cout<<"YES"<<endl;
}
else
{
cout<<"NO"<<endl;
}
}
}
return 0;
}```

Program: Alternate Additions Solution in Java

```import java.util.*;
public class Main
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t--!=0)
{
int a=sc.nextInt();
int b=sc.nextInt();
if(Math.abs(b-a)%3!=1 && Math.abs(b-a)%3!=0)
System.out.println("NO");
else
System.out.println("YES");
}

}
}```

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