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;
typedef  long long int ll;
#define fastttt std::ios::sync_with_stdio(false);std::cin.tie(NULL);std::cout.tie(NULL);

class solver{
	public:
		
	void solve()
	{
		ll n,k;cin>>n>>k;
		
		ll g = k*(n/k);
		ll m=0;
		ll y;
		ll x;
		while(g>0){
			y = sqrt((n*n)-(g*g));
			x = y/k;
			m = max( m , x+(g/k));
			g-=k;
		}
		if(m%2==0)cout<<"Utkarsh\n";else cout<<"Ashish\n";
	}
	
};
 
int main(){
	fastttt
	//	cout<<fixed<<setprecision(10);
	
	int test=1,c=1;
	cin>>test;
	while(test--){
		//cout<<"Case #"<<c<<": ";c++;
		solver o;
		o.solve();
	}
}