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;
using ll = long long;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    ll d, k;
    cin >> d >> k;
    auto lose = [&](ll x, ll y) {
      return (x + k) * (x + k) + y * y > d * d &&
             x * x + (y + k) * (y + k) > d * d;
    };
    ll lb = 0, ub = d + 1;
    while (ub - lb > 1) {
      ll m = (lb + ub) / 2;
      ll x = m * k;
      if (2 * x * x <= d * d)
        lb = m;
      else
        ub = m;
    }
    ll x = lb * k;
    bool win = false;
    if ((x + k) * (x + k) + x * x <= d * d) win = true;
    cout << (win ? "Ashish" : "Utkarsh") << "\n";
  }
}
