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 <iostream>
#include <algorithm>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <vector>
#include <deque>
#include <fstream>
#include <cmath>
using namespace std;


#define MOD 1000000007

int main() {
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin >> t;
	while (t--) {
		int64_t d, k;
		cin >> d >> k;
		int steps = k / d;
		int64_t limit = d * d / (k * k);
		vector<int64_t> a(d+2);
		for (int64_t i = 0; i*i <= limit; i++) {
			if (limit - i * i >= 0)
				a[i] = floor(sqrt((double)limit - i * i));
			else break;
		}
		//vector<pair<int, int> > res;
		int64_t mini = 1000000000;
		for (int i = 0; i <= d; i++) {
			if (a[i] != a[i + 1]) {
				mini = min(abs(i - a[i]), mini);
			}
		}
		//pair<int, int> result;
		//for (int i = 0; i < res.size(); i++) {
		//	if ((res[i].first - res[i].second) < 0) break;
		//	//cout << res[i].first << " " << res[i].second <<" " << (res[i].first - res[i].second) << " " << (((res[i].first + res[i].second) % 2 == 0) ? "utkarsh" : "Ashish") << endl;
		//}

		if (mini % 2 == 0)
			cout << "Utkarsh" << endl;
		else
			cout << "Ashish" << endl;
	}


	return 0;
}
