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;
typedef long long ll;
#define IOS                      \
    ios::sync_with_stdio(false); \
    cin.tie(0)

// *start on @date: 2020-11-22 15:38
// *writer: wjm

int main()
{
    IOS;
    ll t;
    cin >> t;
    while (t--)
    {
        ll k, d;
        cin >> k >> d;
        ll x = d, y = 0;
        ll t = x * x + y * y;
        ll ans = 1;
        while (t <= k * k)
        {
            if (x > y)
            {
                y = y + d;
                ans++;
            }
            else
            {
                x = x + d;
                ans++;
            }
            t = x * x + y * y;
        }
        if (ans % 2 == 0)
        {
            cout << "Ashish"
                 << "\n";
        }
        else
        {
            cout << "Utkarsh"
                 << "\n";
        }
    }
    return 0;
}