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>
#include <unordered_map>

using namespace std;

typedef long long LL;
typedef pair<double, double> P;

const double pi = acos(-1);
const int INF = 0x3f3f3f3f;
const double eps = 1e-10;
const int MAX = 2000;
const int mod = 1e9 + 7;

int main() {
	//freopen("file.in", "r", stdin);
	//freopen("file.out", "w", stdout);
	int T; scanf("%d", &T);
	while (T--) {
		LL a, b; scanf("%lld %lld", &a, &b);
		a = a * a;
		LL k = 0;
		while (2 * k * k <= a) k += b;
		k -= b;
		if ((k + b) * (k + b) + k * k <= a) printf("Ashish\n");
		else printf("Utkarsh\n");
	}
	return 0;
}