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;

#ifdef Dhiraj
    #include "D:/dhiraj/Programming/debug.h"
#else
    #define d(...) 11
    #define cerr if(0) cerr
#endif

#define ll        long long int
#define all(x)    (x).begin(), (x).end()
#define rall(x)   (x).rbegin(), (x).rend()
#define endl      '\n'

template <typename T, typename U>
inline istream& operator >> (istream& in, pair<T, U>& a) {
    in >> a.first >> a.second;
    return in;
}
template <typename T>
inline istream& operator >> (istream& in, vector<T>& a) {
    for(auto& x : a) in >> x;
    return in;
}

void solve(int &T)
{
    ll d, k;
    cin >> d >> k;

    ll x = 0, y = 0, cnt = 0;
    while(x * x + y * y <= d * d) {
        if(x <= y) x += k;
        else y += k;
        cnt++;
    }

    if(cnt % 2 == 0) cout << "Ashish\n";
    else cout << "Utkarsh\n";
}

int main()
{
    #ifdef Dhiraj
        freopen("D:/dhiraj/Programming/i1.txt", "r", stdin);
        freopen("D:/dhiraj/Programming/o1.txt", "w", stdout);
        // freopen("D:/dhiraj/Programming/e1.txt", "w", stderr);
    #endif

    ios::sync_with_stdio(false);
    cin.tie(0);

    int tc = 1;
    cin >> tc;
    for(int i = 1; i <= tc; i++) {
        cerr << "Case #" << i << "\n";
        solve(i);
    }
    return 0;
}