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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long d, k;
    cin >> d >> k;
    long long mx = 0, cur = k, moves = 1;
    while (true) {
      long long md = cur / 2;
      if (cur % 2 == 0) {
        if (md * md + md * md > d * d) {
          break;
        }
      } else {
        if (md * md + (md + 1) * (md + 1) > d * d) {
          break;
        }
      }
      cur += k;
      ++moves;
    }
    --moves;
    if (moves % 2 == 0) {
      long long mn = 0, mx = 0;
      bool ash = true;
      while (true) {
        mx += k;
        if (mn * mn + mx * mx > d * d) {
          ash = false;
          break;
        }
        mn += k;
        if (mn * mn + mx * mx > d * d) {
          ash = true;
          break;
        }
      }
      assert(!ash);
      cout << (ash ? "Ashish\n" : "Utkarsh\n");
    } else {
      long long mn = k / 2, mx = k / 2;
      if (k % 2 == 1) ++mx;
      bool ash = true;
      while (true) {
        mx += k;
        if (mn * mn + mx * mx > d * d) {
          ash = true;
          break;
        }
        mn += k;
        if (mn * mn + mx * mx > d * d) {
          ash = false;
          break;
        }
      }
      cout << (ash ? "Ashish\n" : "Utkarsh\n");
    }
  }
}
