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>
//----------Bajrang Bali ki jai-------------------
//----------Code submitted by Akash Kumar---------
//----------Toh chaliye shuru karte hain----------
#include<bits/stdc++.h>
typedef long long int ll;
#define jaldi       ios::sync_with_stdio(0); cin.tie(0)
#define testcase    ll t;cin>>t;while(t--)
#define fw(i,a,b)   for(i=a;i<b;i++)
#define fw1(i,a,b)  for(i=a;i<=b;i++)
#define bw(i,a,b)   for(i=a;i>b;i--)
#define bw1(i,a,b)  for(i=a;i>=b;i--)
#define pb          push_back
#define vec         vector<ll>
#define mll         map<ll,ll>
#define F           first
#define S           second
#define lcm(a,b)    (a*b)/(__gcd(a,b))
#define yes         cout<<"YES"<<endl
#define no          cout<<"NO"<<endl
#define prt(x)      cout<<x
#define prts(x)     cout<<x<<" "
#define prtl(x)     cout<<x<<endl
#define prtnl()     cout<<"\n"
#define endl        "\n"

const ll mod=1000000007; //10^9+7

using namespace std;

void toh_chaliye_shuru_karte_hain()
{
  #ifndef ONLINE_JUDGE
            freopen("input.txt","r",stdin);
            freopen("output.txt","w",stdout);
    #endif
}

void code_samapti_ki_ghoshna()
{
  #ifndef ONLINE_JUDGE
        cout<<"\nTime Elapsed : " << 1.0*clock() / CLOCKS_PER_SEC << " s\n";
    #endif
}

ll power(ll a,ll b,ll modi)
{
    a%=modi;
    ll res = 1 ;
    while(b)
    {
        if(b%2)
        {
            res = (res * a) % modi ;
        }
        b/=2 ;
        a = (a*a) % modi ;
    }
    return res ;
}

int main()
{
       toh_chaliye_shuru_karte_hain(); //for input and output
       jaldi; // fast input
       testcase
       {
        ll d,k;
        cin>>d>>k;
        ll x=0,y=0;
        while(1)
        {
          x+=k;
          if(x*x+y*y>d*d)
          {
            cout<<"Utkarsh\n";
            break;
          }
          y+=k;
          if(x*x+y*y>d*d)
          {
            cout<<"Ashish\n";
            break;
          }
        }
       }
       code_samapti_ki_ghoshna();
return 0;
}