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<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;
typedef long long ll;
 
const int INF = 0x3f3f3f3f;
ll cal(ll x, ll y) {
	return x * x + y * y;
}
bool check(ll x, ll y, ll d) {
	return cal(x,y) <= d * d;
}
int main()
{
	int t;
	cin >> t;
	while (t--) {
		ll d, k;
		cin >> d >> k;
		ll now = 0;
		ll cnt = 0;
		ll x = 0, y = 0;
		while (now <= d * d) {
			x += k;
			y += k;
			now = cal(x, y);
			cnt++;
		}
		now -= k * k * 2;
		cnt--;
		if (check(cnt * k + k, cnt * k, d)) {
			cout << "Ashish\n";
		}
		else {
			cout << "Utkarsh\n";
		}
	}
	return 0;
}