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 PB push_back
#define f first
#define s second
#define what_is(x) cerr << #x << " is " << x << endl;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;

const int INF = 1000000007;
const ll MOD = 1000000007;

string first = "Ashish";
string second = "Utkarsh";

string find(ll a)
{
    int ctr = 0;
    ll t = 1;
    ll inc = 1;
    while(t < a)
    {
        if(ctr%2 == 0)
        {
            inc+=2;
        }
        t+=inc;
        ctr^=1;
    }
    if(ctr == 0)
    {
        return first;
    }
    else
    {
        return second;
    }
}

void solve_test()
{
    ll d, k;
    cin >> d >> k;
    cout << find((d*d)/(k*k)) << "\n";
}

signed main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    int tests;
    //tests = 1;
    cin >> tests;
    while(tests--)
    {
        solve_test();
    }
    return 0;
}
