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>
//...START BY DOING WHAT IS NECESSARY, THEN WHAT IS POSSIBLE AND SUDDENLY YOU ARE DOING THE IMPOSSIBLE... 
#include <bits/stdc++.h>
using namespace std;
#define FAST_FURIER ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define pb push_back
#define mk make_pair
#define rep(i,a,N) for(ll i=a;i<N;i++)
#define rrep(i,a,N) for(ll i=a;i>N;i--)
typedef long long ll;
#define M 1000000007
bool comp(int x,int y)
{
    return x > y;
}
 
 
/*..............................code starts here........................*/
// C is first won in M
 
 
int main() {
    FAST_FURIER;
    int tt=1;
    cin >> tt;
    ll m,d,k;
   //ll a,b,c;
    while(tt--)
    {
        cin >> d >> k;
        ll x = 0,y = 0;
        int f = 0;
        while(true){
            ll d1 = pow(x+k,2)+pow(y,2), d2 = pow(x,2)+pow(y+k,2);
            f = abs(f-1);
            if(d1 < d2 and d1 <= d*d){
                x += k;
            }
            else if(d2 <= d1 and d2 <= d*d) y += k;
            else break;
        }
        if(f) cout << "Utkarsh";
        else cout << "Ashish";
        cout << endl;

    }
}
 
 
 
// Author : shivam_123
 
// g++ -std=c++17