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.*;
import java.math.*;
import java.io.*;

public class B{
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while(t-- != 0){
			// int n = in.nextInt();
			int d = in.nextInt(), k = in.nextInt();
			int x = k;	
			double dis = x * Math.sqrt(2);
			// boolean bool = false;
			while(dis < d){
				// bool = true;
				x += k;
				dis = x * Math.sqrt(2);
			}
			// if(bool) x -= k;
			x -= k;
			int y = x + k;
			dis = Math.sqrt(x *x + y*y); 
			if(dis <= d){
				System.out.println("Ashish");
			}else{
				System.out.println("Utkarsh");
			}

			// System.out.println(res);
		}
		

		// System.out.println(Arrays.toString(dp));
	
	}
	
}
