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>
using namespace std;
#include<bits/stdc++.h>
#define int long long
#define all(x) x.begin(), x.end()
#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

//DEBUGGER
#define debug1(a) cout<<#a<<" = "<<(a)<<endl;
#define debug2(a,b) cout<<#a<<" = "<<(a)<<", "<<#b<<" = "<<(b)<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<(a)<<", "<<#b<<" = "<<(b)<<", "<<#c<<" = "<<(c)<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<(a)<<", "<<#b<<" = "<<(b)<<", "<<#c<<" = "<<(c)<<", "<<#d<<" = "<<(d)<<endl;

template<class T> ostream& operator<<(ostream& cout,vector<T> V)
{
    cout<<"[ ";for(auto v:V)cout<<v<<' ';return cout<<']';
}

template<class L,class R> ostream& operator<<(ostream& cout,pair<L, R> P)
{
    return cout<<'('<<P.first<<','<< P.second<<')';
}


const int N = 2e3+3;

int d,k;
double buf;
int targ;

int solve()
{
	int ans = 0;
	for ( int i = 0;  ;++i )
	{
		if ( i*i>targ )
			break;

		ans = max(ans, (int)(sqrt( targ - i*i ) + i) );
	}

	return ans;
}

signed main()
{
    IOS;
	int t; cin>>t; while(t--)
	{
		cin>>d>>k;
		
		buf = ((double)d/(double)k);
		buf*=buf;
		targ = floor(buf);

		//debug2(targ,solve());

		if ( solve()%2==0 )
			cout<<"Utkarsh\n";

		else
			cout<<"Ashish\n";
	}    
}