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 <iostream>

using namespace std;

int main() {
#ifndef LOCAL
  cin.tie(0);
#endif
  ios::sync_with_stdio(false);
  int tt;
  cin >> tt;
  while (tt--) {
    long long d;
    int k;
    cin >> d >> k;
    bool can = false;
    {
      int t = 0;
      long long a = 0, b = 0;
      while (a * a + b * b <= d * d) {
        if (t % 2 == 0) {
          a += k;
        } else {
          b += k;
        }
        ++t;
      }
      if ((t - 1) % 2 == 0) {
        can = true;
      }
    }
    {
      int t = 0;
      long long a = 0, b = 0;
      while (a * a + b * b <= d * d) {
        if (t == 0 || t == 1) {
          a += k;
        } else
        if (t % 2 == 0) {
          b += k;
        } else {
          a += k;
        }
        ++t;
      }
      if ((t - 1) % 2 == 0) {
        can = true;
      }
    }
    cout << (can ? "Utkarsh" : "Ashish") << '\n';
  }
}
