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 Main{
	static int[] x = new int[(1 << 16)];
    public static void main(String[] args) throws IOException{
    	In in = new In();
    	for(int qr = in.pint(); qr > 0; qr--) {
    		in.tok();
    		long d = in.plng(), k = in.plng();
    		long z = 0;
    		for(long i = 0; i <= d; i++) {
    			if(i * k * i * k * 2 <= d * d) {
    				z = i;
    			}
    		}
    		long p = z * k;
    		if((p * p) + ((p + k) * (p + k)) <= d * d) {
    			System.out.println("Ashish");
    		}else {
    			System.out.println("Utkarsh");
    		}
    	}
    }
}   
class In{
    BufferedReader in;
    StringTokenizer st = new StringTokenizer("");
    public In(){
        in = new BufferedReader(new InputStreamReader(System.in));
    }
    
    int pint() throws IOException{
        if(st.hasMoreTokens()) {return Integer.parseInt(st.nextToken());}
        else {return Integer.parseInt(in.readLine());}
    } 
    double pdbl() throws IOException{
        if(st.hasMoreTokens()) {return Double.parseDouble(st.nextToken());}
        else {return Double.parseDouble(in.readLine());}
    }
    long plng() throws IOException{
        if(st.hasMoreTokens()) {return Long.parseLong(st.nextToken());}
        else {return Long.parseLong(in.readLine());}
    }
    char pchr() throws IOException{
        if(st.hasMoreTokens()) {return st.nextToken().charAt(0);}
        else {return in.readLine().charAt(0);}
    }
    String pstr() throws IOException{
        if(st.hasMoreTokens()) {return st.nextToken();}
        else {return in.readLine();}
    }
    String readLine() throws IOException{
        return in.readLine();
    }
    boolean ready() throws IOException {return in.ready();}
    boolean readyN() throws IOException{return in.ready() || st.hasMoreTokens();}
    boolean tokens() {return st.hasMoreTokens();}
    void tok() throws IOException{st = new StringTokenizer(in.readLine());}
    void skip() throws IOException{in.readLine();}
}