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<iostream>
#include<cstdio>
#include<cmath>
#define int long long
using namespace std;
inline int read()
{
	int n=0,f=1,ch=getchar();
	while(ch<'0'||ch>'9')
	{
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9')
	{
		n=n*10+ch-'0';
		ch=getchar();
	}
	return n*f;
}
signed main()
{
	int t,d,k;
	bool flag;
	t=read();
	for(int greg=1;greg<=t;greg++)
	{
		d=read();
		k=read();
		flag=false;
		for(int i=0;i<=d/k;i++)
		{
			if((long long)(sqrt((d*d)/(k*k)-i*i))==i)
			{
				flag=true;
				break;
			}
		}
		if(flag==false)printf("Ashish\n");
		else printf("Utkarsh\n");
	}
	return 0;
}
