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>
#include<iostream>
using namespace std;
#define int long long int
#define ld long double
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define removeDuplicates(a) a.resize(unique(all(a))-a.begin())
#define endl '\n'	
#define setbits(x) (__builtin_popcount(x))
int mod=1000000007;
int INF64=mod*mod+1;

void solve(){
	int d,k;
	cin>>d>>k;

	int x;
	for(x=0;2*x*x<=d*d;x+=k);

	x-=k;
	if(x*x+(x+k)*(x+k)<=d*d)
		cout<<"Ashish"<<endl;
	else
		cout<<"Utkarsh"<<endl;
}

int32_t main(){
	
	ios_base::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);

	int T=1;
	cin>>T;
	for(int i=1;i<=T;i++){
		// cout<<"Case #"<<i<<": ";
		solve();
	}
	cerr<<"Time : "<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n";
}