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 LOCAL
    #define D(a) cerr << #a << " = " << a << endl
#else 
    #define D(a) 8 
#endif
#define fastio ios_base::sync_with_stdio(0); cin.tie(0)
#define dforsn(i,s,n) for(int i=int(n-1);i>=int(s);i--)
#define forsn(i,s,n) for(int i=int(s);i<int(n);i++)
#define all(a) (a).begin(),(a).end()
#define dforn(i,n) dforsn(i,0,n)
#define forn(i,n) forsn(i,0,n)
#define si(a) int((a).size())
#define pb emplace_back
#define mp make_pair
#define snd second
#define fst first
#define endl '\n'
using pii = pair<int,int>;
using vi = vector<int>;
using ll = long long;

int main() {
	fastio;
	
    int t; cin >> t;
    while (t--) {
        int d, k; cin >> d >> k;

        ll k2 = ll(k) * k, d2 = ll(d) * d, q = d2 / k2;
        int l = 0, h = d;
        while (h-l > 1) {
            int m = (l + h) / 2;
            if (2LL*m*m <= q) l = m;
            else h = m;
        }
        bool first_wins = 2LL*l*l + 2*l + 1 <= q;
        cout << (first_wins ? "Ashish" : "Utkarsh") << endl;
    }
	
	return 0;
}
