Nastia has 2 positive integers A and B. She defines that:

  * The integer is good if it is divisible by A ⋅ B; 
  * Otherwise, the integer is nearly good, if it is divisible by A. 



For example, if A = 6 and B = 4, the integers 24 and 72 are good, the integers 6, 660 and 12 are nearly good, the integers 16, 7 are neither good nor nearly good.

Find 3 different positive integers x, y, and z such that exactly one of them is good and the other 2 are nearly good, and x + y = z.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains two integers A and B (1 ≤ A ≤ 10^6, 1 ≤ B ≤ 10^6) — numbers that Nastia has.

Output

For each test case print: 

  * "YES" and 3 different positive integers x, y, and z (1 ≤ x, y, z ≤ 10^{18}) such that exactly one of them is good and the other 2 are nearly good, and x + y = z. 
  * "NO" if no answer exists. 

You can print each character of "YES" or "NO" in any case.

If there are multiple answers, print any.

Example

Input


3
5 3
13 2
7 11


Output


YES
10 50 60
YES
169 39 208
YES
28 154 182

Note

In the first test case: 60 — good number; 10 and 50 — nearly good numbers.

In the second test case: 208 — good number; 169 and 39 — nearly good numbers.

In the third test case: 154 — good number; 28 and 182 — nearly good numbers.
#include <bits/stdc++.h>
 
#define ll long long
#define pb push_back
#define rep(i,a,b) for (int i =a; i <b; ++i)
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define mod 1000000007
 
using namespace std;
 
/*bool comp(int a, int b)
{
	return a>b;
}*/
//const ll M=1e9+7;
/*int power(int x, unsigned int y)
{
    if (y == 0)
        return 1;
    else if (y % 2 == 0)
        return power(x, y / 2) * power(x, y / 2);
    else
        return x * power(x, y / 2) * power(x, y / 2);
}*/
 
int main()
{
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);
	cout.tie(NULL);

	/*#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif*/


	int t=1;
	cin>>t;
 
	while(t--)
	{
		ll a,b;
		cin>>a>>b;

	
		ll x,y,z;
		x=a;
		y=a*b;
		z=x+y;

		if(b==1)
			no;
		else
		{
			yes;
			cout<<x<<" "<<y<<" "<<z<<endl;
		}
 		
	}
	
	return 0;
}