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;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int t;
	cin >> t;
	while (t-- > 0) {
		int d, k;
		cin >> d >> k;
		long long sqr = d * 1ll * d / (k * 1ll * k);
		int y = 0;
		while (2 * y * 1ll * y <= sqr) {			
			++y;
		}
		--y;
		int x = 0;
		while (x * 1ll * x + y * 1ll * y <= sqr) {
			++x;
		}
		--x;
		cout << (((x + y) % 2 == 0) ? "Utkarsh" : "Ashish") << endl;
	}
	return 0;
}