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 <iostream>
#include <vector>
#include <string>
#include <set>
#include <algorithm>
#include <map>

using namespace std;

int main() {
    int t;
    cin>>t;
    for (int re=0; re<t; re++) {
        long long d, k;
         cin>>d>>k;
         long long A=0;
         for (long long a=0; a<=d+1; a++) {
             if (2*a*a*k*k>d*d) {
                 A=a-1;
                 break;
             }
         }
         long long nline=2*A;
         if (A*A*k*k+(A+1)*(A+1)*k*k<=d*d) {
             nline++;
         }
         if (nline%2==0) {
             cout<<"Utkarsh"<<endl;
         } else cout<<"Ashish"<<endl;

    }
    return 0;
}
