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<algorithm>
#include<map>
#include<stack>
#include <stdio.h>
#include <string.h>
#include<iomanip>
#include<random>
#include<set>
#include<fstream>
#include<unordered_set>
#include<unordered_map>
#include<cmath>
#include<queue>
#include<immintrin.h>
using namespace std;
#define ll long long
#define int long long
bool ok(ll x, ll y, ll d)
{
    return x*x+y*y<=d*d;
}
signed main() {
    std::ios_base::sync_with_stdio(0);
    cin.tie(0);
    int q;
    cin>>q;
    while(q--)
    {
        ll d, k;
        cin>>d>>k;
        ll z=0;
        while(ok(k*(z+1), k*(z+1), d))
            z++;
        if(!ok(k*z, k*(z+1), d))
            cout<<"Utkarsh";
        else
            cout<<"Ashish";
        cout<<'\n';
    }
    return 0;
}