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 pb          push_back
#define pii         pair<int,int>
#define vi          vector<int>
#define vii         vector<pii>
#define mi          map<int,int>
#define mii         map<pii,int>
#define all(a)      (a).begin(),(a).end()
#define f           first
#define s          second
#define sz(x)       (int)x.size()
#define rep(i,a,b)  for(int i=a;i<b;i++)
#define rep0(a,n)  for(int i=0;i<n;i++)cin>>a[i]
#define si(x)       cin>>x
#define sii(x,y)    cin>>x>>y
#define siii(x,y,z) cin>>x>>y>>z
#define ahell   998244353
#define hell 1000000007
#define N1 100001
#define fl 1.0000000000
//#define mod 998244353
#define cntbit(a) __builtin_popcountll(a)
#define ios ios_base::sync_with_stdio(false); cin.tie(NULL);
using namespace std;

int n,m,k;
vi adj[2*N1];

main()
{
  ios
  string str;
  int i,j,d;
  si(n);
  while(n--)
  {
    sii(d,k);
    int t=d/(2*k);
    int flg=0;
    for(j=max(0LL,t-100);j<=t+100;j++)
    {
        if(((2*j*j*k*k)<=(d*d))&&((j*j*k*k+(j+1)*(j+1)*k*k)>d*d))
            {
                flg=2;
            }
        else if(((j*j*k*k+(j+1)*(j+1)*k*k)<=d*d)&&((2*(j+1)*(j+1)*k*k)>d*d))
        {
            flg=1;
        }
        if(flg)break;
    }
    if(flg==1)
    {
        cout<<"Ashish";
    }
    else cout<<"Utkarsh";
    cout<<"\n";
}
}



