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.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class problem4 {
    public static void main(String[] args) throws Exception{
        var reader = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(reader.readLine());
        for (int i = 0; i < t; i++) {
            var tokenizer = new StringTokenizer(reader.readLine());
            long d = Integer.parseInt(tokenizer.nextToken());
            long k = Integer.parseInt(tokenizer.nextToken());
            int odd = 0; // нечётный
            int even = 0; // чётный
            boolean winner = true;
            for (long x = 0; x <= d; x+=k) {
                long y = (long)Math.sqrt(d*d - x*x);
                y -= (y % k);
                if(((x + y)/k) % 2 == 0) even++;
                else odd++;
                if(x == y) winner = false;
            }
            if(winner || even == 0) System.out.println("Ashish");
            else System.out.println("Utkarsh");
        }
    }
}
