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>
/****************************************

	Author: LostMartian (Sahil Jitesh Gangurde)

****************************************/
//#pragma GCC optimize("Ofast")  
//#pragma GCC target("avx,avx2,fma") 
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void pn(Arg1&& arg1) { cout << arg1 << "\n"; }
template <typename Arg1, typename... Args>
void pn(Arg1&& arg1, Args&&... args) { cout << arg1 << "\n"; pn(args...); }
template <typename Arg1>
void ps(Arg1&& arg1) { cout << arg1 << " "; }
template <typename Arg1, typename... Args>
void ps(Arg1&& arg1, Args&&... args) { cout << arg1 << " "; ps(args...); }
template <typename Arg1>
void rn(Arg1&& arg1) { cin >> arg1; }
template <typename Arg1, typename... Args>
void rn(Arg1&& arg1, Args&&... args) { cin >> arg1; rn(args...); }
template <typename T>
inline T GCD(T a, T b) { while (b != 0)swap(b, a %= b); return a; }
template <typename T>
T binpow(T a, T b) {T res = (T)1; if (b == (T)0) {return (T)1;} while (b > (T)0) {if (b & (T)1) {res *= a;} a *= a; b /= (T)2;} return res;}
#define d(x) cout<<"Value of "#x" = "<<x<<"\n";
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#define iAmSpeed ios_base::sync_with_stdio(false); cin.tie(NULL);
#define LOCAL freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define ll long long
#define rep(i,m,n) for(ll i=m; i<n; i++)
#define repr(i,m,n) for(ll i=m; i>=n; i--)
#define vll vector<ll >
#define vvl vector<vector<ll > >
#define pi pair<ll, ll >
#define vpi vector<pi >
#define umap unordered_map<ll, ll >
#define pb push_back
#define mk make_pair
#define sz size()
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define test ll t; cin>>t; while(t--)
#define ini(a, x) memset(a, x, sizeof(a))
void solve() {
	ll d, k;
	cin>>d>>k;
	ll divs=(d/k)*k;
	ll y=divs;
	ll cnt=0;
	while(true) {
		if((divs*divs)+(y*y)<=(d*d)) {
			break;
		}
		else {
			if(cnt%2==0) { 
				divs-=k;
			}
			else { 
				y-=k;
			}
			cnt++;
		}
	}
	ll ans=(divs/k) + (y/k);
	cout<<(ans&1?"Ashish\n":"Utkarsh\n");
}
int main()
{
	clock_t z = clock();
	//LOCAL();
	iAmSpeed
	test {
		solve();
	}









	debug("Total Time: %.7f\n", (double)(clock() - z) / CLOCKS_PER_SEC);
	return 0;
}
