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>

#define pb push_back
#define mp make_pair
#define f first
#define s second


#define all(x) (x).begin(), (x).end()
#define sz(a) (int)(a).size()
#define for0(i, n) for (int i = 0; i < (int)(n); i++)
#define for1(i, n) for (int i = 1; i <= (int)(n); i++)
#define int long long

using namespace std;
typedef pair<int, int> pii;
typedef long double ld;


/*
    ID:forelise
    TASK:comehome
    LANG:C++
*/



signed main(){
   //ifstream cin("input.txt");
  //ofstream cout("comehome.out");
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int u;
cin>>u;
while(u--)
{

int d, k;
cin>>d>>k;

bool ok = 0;

int L = 0, R = 1;

while (2 * R * R * k * k < d * d)
R *= 2;

while(R - L > 1)
{
    int M = (R + L) / 2;
    if (2 * M * M * k * k <= d * d)
        L = M;
    else
        R = M;
}

int ss = L * k;



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


}


}










