Utkarsh is forced to play yet another one of Ashish's games. The game progresses turn by turn and as usual, Ashish moves first.

Consider the 2D plane. There is a token which is initially at (0,0). In one move a player must increase either the x coordinate or the y coordinate of the token by exactly k. In doing so, the player must ensure that the token stays within a (Euclidean) distance d from (0,0).

In other words, if after a move the coordinates of the token are (p,q), then p^2 + q^2 ≤ d^2 must hold.

The game ends when a player is unable to make a move. It can be shown that the game will end in a finite number of moves. If both players play optimally, determine who will win.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.

The only line of each test case contains two space separated integers d (1 ≤ d ≤ 10^5) and k (1 ≤ k ≤ d).

Output

For each test case, if Ashish wins the game, print "Ashish", otherwise print "Utkarsh" (without the quotes).

Example

Input


5
2 1
5 2
10 3
25 4
15441 33


Output


Utkarsh
Ashish
Utkarsh
Utkarsh
Ashish

Note

In the first test case, one possible sequence of moves can be

(0, 0) \xrightarrow{Ashish } (0, 1) \xrightarrow{Utkarsh } (0, 2).

Ashish has no moves left, so Utkarsh wins.

<image>
import java.util.Scanner;
import java.io.*;
import java.lang.*;
import java.util.*;


public class ExB{

    static class IntIntPair{ 
        private int xValue; 
        private int yValue; 
         
        public IntIntPair(int xValue, int yValue){ 
            this.xValue = xValue; 
            this.yValue = yValue; 
        }   
         
        public int getx(){ 
            return this.xValue; 
        } 
         
        public int gety(){ 
            return this.yValue; 
        }   
    }
    
    public static void main(String args[])
    {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        for(int i=0; i<t; i++){
            double d = scan.nextLong();
            double k = scan.nextLong();
            d =  Math.pow(d, 2);
            k =  Math.pow(k, 2);
            long n = (long) Math.pow(d/k*2,0.5);
            if(n%2 == 0 || n == 0)
                System.out.println("Utkarsh");
            else
                System.out.println("Ashish");
        }
        

            
        
    }  

        
}
