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 ll long long
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		ll d,k;
		scanf("%lld %lld",&d,&k);
		ll x = 0LL,y = 0LL,step = 0LL;
		while(true)
		{
			if(step)x+=k;
			else y+=k;
			if(x*x+y*y>d*d)break;
			step^=1;
		}
		if(step==0)
		{
			printf("Utkarsh\n");
		}
		else 
		{
			x = k;
			y = 0;
			step = 1;
			while(true)
			{
				if(step)x += k;
				else y+=k;
				if((x*x+y*y)>d*d)break;
				step^=1;
			}
			if(step==0)
			{
				printf("Utkarsh\n");
			}
			else 
			{
				printf("Ashish\n");
			}
		}
	}
}