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 <algorithm>
#include <cstdio>
#include <vector>
#include <stack>
#include <string>
#include <cstring>
#include <queue>
#include <map>
#include <set>
#include <cmath>
#include <climits>
#include <functional>

typedef long long ll;

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("in1.txt", "r", stdin);
    // freopen("w.txt", "w", stdout);
#endif
    // ios::sync_with_stdio(false);
    
    ll T, d, k;
    for (cin >> T; T--;) {
        cin >> d >> k;

        ll cnt = 0;
        while ((cnt * k) * (cnt * k) * 2 <= d * d) {
            ++cnt;
        }
        --cnt;
        
        ll res = (cnt * k) * (cnt * k) + (cnt * k + k) * (cnt * k + k);

        cout << (res <= d * d ? "Ashish" : "Utkarsh") << endl;

    }
}