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 D {
    public static void main(String args[]) throws Exception {
        BufferedReader br = new BufferedReader (new InputStreamReader (System.in));
        Scanner sc = new Scanner (System.in);

        int T = sc.nextInt();
        // int T = Integer.parseInt(br.readLine());
        while(T-->0){
            int d = sc.nextInt();
            int k = sc.nextInt();
            long arr[] = new long[d];
            int pos = 0;
            for(int i=1; (long)(k)*k*i*i<(long)d*d; i++){
                arr[pos] = (long)(k)*((long)k*i*i);
                pos++;
            }
            long max1 = 0, max2 = 0;
            for(int i=pos-1; i>=0; i--){
                if(i<0) break;
                for(int j=i-1; j>=0; j--){
                    if(arr[i] + arr[j]<=(long)d*d && (Math.sqrt(max1) + Math.sqrt(max2)<Math.sqrt(arr[i]) +Math.sqrt(arr[j]))){
                        max1 = arr[i];
                        max2 = arr[j];
                    }
                }
            }
            int ans = (int)(Math.sqrt(max1) + Math.sqrt(max2)) / k;
            if(ans%2 == 1){
                System.out.println("Ashish");
            }else{
                System.out.println("Utkarsh");
            }
        }
    }
}
