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.io.PrintWriter;
import java.util.Scanner;

public class Main {
    
    public static void main(String[] args) {
        
        var sc = new Scanner(System.in);
        var pw = new PrintWriter(System.out);
        
        int T = Integer.parseInt(sc.next());
        for(int t = 0; t < T; t++){
            long d = Long.parseLong(sc.next());
            long k = Long.parseLong(sc.next());
            
            long low = 0L;
            long high = 1000000000L;
    
            while(high - low > 1){
                long mid = (low + high) / 2;
                if(ok(mid, d)){
                    low = mid;
                }else{
                    high = mid;
                }
            }
            
            long a = low - low % k;
            if(a*a + (a+k)*(a+k) <= d*d){
                pw.println("Ashish");
            }else{
                pw.println("Utkarsh");
            }
        }
        pw.flush();
    }
    
    static boolean ok(long mid, long d){
        
        return mid*mid + mid*mid <= d*d;
    }
}