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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    long long d, k;
    cin >> d >> k;
    long long s = d;
    vector<string> ret(2);
    ret[0] = "Utkarsh";
    ret[1] = "Ashish";
    bool finish = false;
    for (auto i = 0; i <= d; ++i) {
      for (auto j = s; j >= 0; --j) {
        if (j * j * k * k + (j + i) * (j + i) * k * k <= d * d) {
          if (j * j * k * k + (j + i + 1) * (j + i + 1) * k * k > d * d) {
            cout << ret[i % 2] << endl;
            finish = true;
            break;
          }
          s = j;
          break;
        }
      }
      if (finish) {
        break;
      }
    }
  }
}
