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<bits/stdc++.h>

using namespace std;

#define fi first
#define se second
#define pb push_back
#define ins insert
#define fs(n) fixed<<setprecision(n)
#define F(i, n) for(int i = 0; i < n; i++)

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef map<int, int> mii;
typedef pair<int, int> pii;
typedef set<int> si;
typedef multiset<int> msi;

const int MOD = 1e9 + 7;

void solve() {
	ll d, k;
	cin >> d >> k;
	
	ll x = d / sqrt(2);
	x = x - x % k;
	
	if (x * x + (x + k) * (x + k) <= d * d) cout << "Ashish\n";
	else cout << "Utkarsh\n";
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);

//	freopen("1.txt", "r", stdin);
//	freopen("2.txt", "w", stdout);

	int t;
	cin>>t;

	while(t--)
		solve();

	return 0;
}


