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_base::sync_with_stdio(0);
  cin.tie(0);
  std::cout.tie(0);
  long long t, q, x, l, y, r, g, k, c, m, p;
  long long n;
  string s;
  cin >> t;
  vector<long long> a(100000);
  for (long long i = 1; i <= 100000; i++) {
    a[i - 1] = long(i) * i;
  }
  while (t--) {
    double d;
    cin >> d >> k;
    long double track = d / k;
    track = pow(track, 2);
    int temp = 0;
    for (int i = 0; i < 100000; i++) {
      if (a[i] > track / 2) {
        temp = i;
        break;
      }
    }
    int temp2 = 0;
    track = track - a[temp];
    for (int i = 0; i < 100000; i++) {
      if (a[i] > track) {
        temp2 = i - 1;
        break;
      }
    }
    if ((temp + temp2) % 2 == 0) {
      cout << "Utkarsh" << endl;
    } else {
      cout << "Ashish" << endl;
    }
  }
  return 0;
}
