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 int
#define pb push_back
#define mp make_pair
#define up upper_bound
#define lp lower_bound
#define vl vector<ll>::iterator
using namespace std;
void fast()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

     std::ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}
int main()
{
 fast();
 ll t;
 cin>>t;
 while(t--)
 {
     ll a,b;
     cin>>a>>b;
     ll x=a*b;
     ll y=a;
     if(b==1)
     {
         cout<<"NO"<<endl;
     }
     else
     {
     while(1)
     {
         if(y%x==0||(y+x)%x==0)
         y=y+a;
         else
         break;
         
 }
 cout<<"YES"<<endl;
 cout<<x<<" "<<y<<" "<<x+y<<endl;
}
}
}