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>
#include<bits/stdc++.h>
using namespace std ;
#define pb push_back
#define mem( ar, val) memset( ar , val , sizeof( ar) ) 
typedef long long int lli ;   

lli rec( lli x , lli y , lli r , lli chance ) 
{
	lli temp  = x * x + y * y  ;
	if( temp > r * r )  return 0 ; 
	return 0 ; 
} 


int main()
{
	lli t;
	cin >> t ;
	while( t-- )
	{
		lli n , k ;
		cin >> n >> k ; 
		lli temp = n * n ; 

		lli x = 0 , y = 0 ;
		lli ans = 0 ; 
		while( 1 ) 
		{
			lli temp  = x * x + y * y  ;
			if( temp > n * n ) break ; 
			if( ans ) x += k ;
			else y += k ;
			ans ^= 1 ;

		}

		if( !ans ) cout << "Ashish\n" ;
		else cout << "Utkarsh\n" ;
	}
}