## Algorithm

A. Add Odd or Subtract Even
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given two positive integers a and b.

In one move, you can change a in the following way:

• Choose any positive odd integer x (x>0�>0) and replace a with a+x�+�;
• choose any positive even integer y (y>0�>0) and replace a with ay�−�.

You can perform as many such operations as you want. You can choose the same numbers x and y in different moves.

Your task is to find the minimum number of moves required to obtain b from a. It is guaranteed that you can always obtain b from a.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1t1041≤�≤104) — the number of test cases.

Then t test cases follow. Each test case is given as two space-separated integers a and b (1a,b1091≤�,�≤109).

Output

For each test case, print the answer — the minimum number of moves required to obtain b from a if you can perform any number of moves described in the problem statement. It is guaranteed that you can always obtain b from a.

Example
input
Copy
5
2 3
10 10
2 4
7 4
9 3

output
Copy
1
0
2
2
1

Note

In the first test case, you can just add 11.

In the second test case, you don't need to do anything.

In the third test case, you can add 11 two times.

In the fourth test case, you can subtract 44 and add 11.

In the fifth test case, you can just subtract 66.

## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

<#include<bits/stdc++.h>
using namespace std;

int main(){
int t;
cin>>t;
while(t--){
int a, b;
cin>>a>>b;
if(a > b){
if((a%2 == 0 && b%2 == 0) || (a%2 != 0 && b%2 != 0)){
cout<<1<<endl;
continue;
}
cout<<2<<endl;
continue;
}
if(a <<b){
if((a%2 == 0 && b%2 == 0) || (a%2 != 0 && b%2 != 0)>{
cout<<2<<endl;
continue;
}
cout<<1<<endl;
continue;
}
cout<<0<<endl;
}
return 0;
}
Copy The Code &

Input

cmd
5
2 3
10 10
2 4
7 4
9 3

Output

cmd
1
0
2
2
1