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 nax = 1e5 + 5;
double eps = 1e-9;
ll mul = 1 << 20;
vector<ll> squares(nax);
unordered_map<ll, bool> m;
ll d, k;
pair<int, int> getxy(ll num) { return {num / mul, num % mul}; }
ll gethash(int x, int y) { return {x * mul + y}; }
bool go(int x, int y, int turn) {
  ll num = gethash(x, y);
  if (m.count(num)) {
    return m[num];
  }
  bool b1 = go(x, y + k, turn ^ 1);
  bool b2 = go(x + k, y, turn ^ 1);
  return m[num] = !(b1 & b2);
}
int main() {
  int t;
  for (int i = 0; i < nax; i++) {
    squares[i] = (ll)i * i;
  }
  cin >> t;
  while (t--) {
    cin >> d >> k;
    int py = nax, summx = 0;
    for (int i = 0; i <= d; i += k) {
      int x = i;
      int y2 = d * d - (ll)x * x;
      int y = lower_bound(squares.begin(), squares.begin() + py, y2) -
              squares.begin();
      if (y2 == squares[y]) {
        y++;
      }
      if (y % k) {
        y += (k - y % k);
      }
      py = y + 1;
      summx = max(summx, (x + y) / (int)k);
    }
    cout << ((summx & 1) ? "Utkarsh\n" : "Ashish\n");
  }
  return 0;
}
