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 pair<int, int> pr;
typedef vector<int> vec;
typedef long long int lli;

#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define TEST int T; cin >> T; while(T--)

#define rep(i,n) for (auto i = 0; i < n; ++i) 
#define repk(i,k,n) for (auto i = k; i < n; ++i) 
#define repr(i,k,n) for (auto i = k; i >= n; --i) 

void solve() {
    lli d, k;
    cin >> d >> k;
    lli p = sqrt((d*d)/(2*k*k));
    lli rem = d*d - 2*p*p*k*k;
    if(rem >= k*k*(2*p+1)) cout << "Ashish\n";
    else cout << "Utkarsh\n";
    // cout << p << ' ' << rem << endl;
}

int main() {
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    TEST {
        solve();
    }
}