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 endl "\n"
#define dbg(x...) do { cout << #x << " -> "; err(x); } while (0)
void err() { cout << endl; }
template <class T, class... Ts>
void err(const T& arg, const Ts&... args) {
    cout << arg << ' '; err(args...);
}

int d, k;

double getDis(int x, int y) {
    return sqrt(1.0 * x * x + 1.0 * y * y);
}

void solve() {
    cin >> d >> k;
    for (int i = 0; ; ++i) {
        if (getDis(i * k + k, i * k + k) > d) {
            if (getDis(i * k, i * k + k) > d) {
                puts("Utkarsh");
            }
            else puts("Ashish");
            return;
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);cout.tie(nullptr);
    int _T = 1;
    cin >> _T;
    while (_T--) solve();
    return 0;
}