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.*; 
import java.util.*; 
import java.math.*; 

public class Main { 
	public static HashMap<Integer,Integer> primeFactors(long n) 
    { 
		HashMap<Integer,Integer> ans=new HashMap<>();
        int temp=0;
        while (n%2==0) 
        { 
            n /= 2; 
            temp+=1;
        } 
        if(temp>0) {
        	ans.put(2,temp);
        }
        
//        System.out.println(Math.sqrt(n));
        for (int i = 3; i <= Math.sqrt(n); i+= 2) 
        { 
            temp=0;
            
            while (n%i == 0) 
            { 
                temp++; 
                n /= i; 
            } 
            if(temp>0) {
            	ans.put(i,temp);
            }
//            System.out.println(n+" "+i);
        } 
        if (n > 2) {
        	ans.put((int)n, 1);
        }
        return ans;	
    } 
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=f.nextInt();
//		int t=1;
		while(t-->0){
			int d=f.nextInt();
			int k=f.nextInt();
			double diagonal=((double)k)*((double)Math.sqrt(2));
			int parts=(int)(((double)d)/diagonal);
//			System.out.println((parts-(int)parts));
			if(((k*parts)*(k*parts)+(k*(parts+1))*(k*(parts+1)))>(d*d)) {
				System.out.println("Utkarsh");
			}
			else {
				System.out.println("Ashish");
			}
		}
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 