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 int long long
#define F(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
#define R(i,b,a) for(int i=(int)(b);i>=(int)(a);i--)
#define endl "\n"
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define out(x) cout<<setprecision(15)<<x<<endl,exit(0)
#define pii pair<int,int>
#define pb push_back
#define all(v) v.begin(),v.end()
#define I first
#define S second
#define ld long double
ld pi=3.14159265359;

int32_t main(){
   ios;
   int t;
   cin>>t;
   while(t--){
   		int d,k;
   		cin>>d>>k;
   		int x=double(d)/sqrt(2)/double(k);
   		x*=k;
   		
   		if((x+k)*(x+k)+x*x<=d*d) cout<<"Ashish"<<endl;
   		else cout<<"Utkarsh"<<endl;
   }
} 
