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>
// Problem: D. Circle Game
// Contest: Codeforces - Codeforces Round #685 (Div. 2)
// URL: https://codeforces.com/contest/1451/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <iostream>
#include <iomanip>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <cctype>
#include <climits>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <stdexcept>
#include <exception>
#include <fstream>
#include <algorithm>
#include <ios>
#include <bitset>
#include <sstream>
#include <set>
#include <list>
#include <unordered_map>
#include <unordered_set>

#define test 1
#define pi acos(-1.0)
#define eps 1e-9
#define MAX(a, b) (((a) > (b)) ? (a) : (b))
#define MIN(a, b) (((a) < (b)) ? (a) : (b))
#define ABS(x) (((x) > 0) ? (x) : (-(x)))
#define FOR(i,value, n) for(LL i = value ; i< n ;i++)
#define FORR(i,value, n) for(LL i = value ; i >= n ;i--)
#define ALL_SET(a, range, value_for) FOR(initial, 0, range)a[initial] = value_for
#define vsize(v) ((LL)v.size())
#define PP(x) printf("%lld\n", (x))
#define CC(x) cout << (x) << '\n'
#define NL cout << '\n'
#define end '\n'
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define F first
#define SE second

using namespace std ;

typedef int I ;
typedef short int SI ;
typedef long L ;
typedef long long LL ;
typedef unsigned long long ULL ;
typedef double D ;
typedef long double LD ;
typedef char C ;
typedef bool B ;
typedef string S ;
typedef vector<I> VI ;
typedef vector<I>::iterator VIT ;
typedef vector<L> VL ;
typedef vector<L>::iterator VLT ;
typedef vector<LL> VLL ;
typedef vector<LL>::iterator VLLT ;
typedef vector<C> VC ;
typedef vector<S> VS;
typedef vector<C>::iterator VCT ;
typedef pair<LL , LL> PLL;
typedef vector<PLL> VPLL;
typedef vector<VPLL> VVPLL;
typedef set<PLL>::iterator SPLLIT;
typedef unordered_map<C, VC> UMCVC;
template<typename a>
void SWAP(a &b, a &c)
{
    a d = b ;
    b = c ;
    c = d ;
}

LL power(LL a, LL b)
{
    LL result = 1 ;
    while (b)
    {
        if (b & 1)
            result = a * result ;
        a *= a ;
        b >>= 1 ;
    }
    return result ;
}
LL gcd(LL a, LL b)
{
    while (b)
    {
        a = a % b;
        SWAP(a, b);
    }
    return a;

}

LL lcm(LL a, LL b)
{
    return (a * (b / gcd(a, b)));
}
LL check_double_equality(D a, D b)
{
    D c = ABS(a-b);
    if(c < eps )
        return true;
    else
        return false;
}
int main()
{
 	   // #ifndef ONLINE_JUDGE
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    // #endif

    ios::sync_with_stdio(0);
    cin.tie(0);
    LL t;
    cin >> t;
    while(t--)
    {
        LL d, k;
        cin >> d >> k;
        LL  ans = 0;
        LL x = 0, y = 0;
		while(1){
			x += k;
			LL rootval = (x*x) + (y*y);
			y += k;
			LL val = (x*x) + (y*y);
			if(val <= (d*d)){
				ans += 2;
			}
			else{
				if(rootval <= (d*d)){
					ans++;
				}
				break;
			}
		}
		// CC(ans);
		if(ans % 2 == 0){
			CC("Utkarsh");
		}
		else{
			CC("Ashish");
		}
    }
    return 0;
}