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.*;

public class Codeforces685 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int i = 0; i<t; i++) {
            System.out.println(solve(sc));
        }
    }

    public static String solve(Scanner sc) {
        int d = sc.nextInt();
        int k = sc.nextInt();
        int p = 0;
        int q = 0;
        boolean ashish = true;

        while (p * p + q * q <= d * d) {
            boolean did = false;
            if (Math.max(p, q) == p) {
                if ((p + k) * (p + k) + q * q <= d * d) {
                    p += k;
                    did = true;
                } else {
                    if (p * p + (q + k) * (q + k) <= d * d) {
                        q += k;
                        did = true;
                    }
                }
            } else {
                if (p * p + (q + k) * (q + k) <= d * d) {
                    q += k;
                    did = true;
                } else {
                    if ((p + k) * (p + k) + q * q <= d * d) {
                        p += k;
                        did = true;
                    }
                }
            }

            if (!did)
                break;
            ashish = !ashish;
        }

            if (!ashish)
                return "Ashish";
            return "Utkarsh";
        }

    }
