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.io.*;

public class Question4 {

	static Scanner sc = new Scanner(System.in);
	public static void main(String[] args) {
		int t = sc.nextInt();
		while(t-->0) {
			solve();
		}
	}
	
	public static void solve() {
		
		//input
		int d = sc.nextInt();
		int k = sc.nextInt();
		
		int a = binary((long)d * d);
		int rem = a % k;
		//System.out.println(a);
		a -= rem;
	
		//System.out.println(a);
		
		long ax = (long)a * a;
		long bx = (long)((long)((long)a + k) * ((long)a + k)); 
		
		ax += bx;
		//System.out.println("ax : " + ax);
	
		
		long dx = (long)d * d;
		//System.out.println("dx : " + dx);
		
		
		if(ax <= dx) {
			System.out.println("Ashish");
		}else System.out.println("Utkarsh");
	}
	
	public static int binary(long d) {
		
		long l = 0, r = d;
		long ans = 0;
		while(l <= r) {
			long mid = (l + r)/2;
			if((long)mid * mid * 2L <= d) {
				ans = mid;
				l = mid + 1;
			}
			else r = mid - 1;
		}
		
		return (int)ans;
	}
}
