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
#define ld long double
#define pi (3.141592653589)
#define pb push_back
#define mp make_pair
#define INF 1000000000
#define MOD 1000000007
#define vll vector<ll>
#define vpll vector<pair<ll,ll> >
#define pll pair<ll>
#define ff first
#define ss second


int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
ll t;
cin>>t;
while(t--){
ll d,k;
cin>>d>>k;
ll j=k;
ll ma=0;
ll a,b;
while(d>=j)
{
   a=j/k;
   b=(d*d)-(j*j);
   b=sqrt(b);
   a+=(b/k);
   if(a>ma)
   ma=a;
   j+=k;
}
if(ma%2)
cout<<"Ashish"<<"\n";
else
cout<<"Utkarsh"<<"\n";
}
return 0;
}
