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;
void eea(long long int a, long long int b, long long int &x, long long int &y) {
  if (a == 1) {
    x = 1;
    y = 0;
    return;
  }
  long long int x1, y1;
  eea(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
}
long long int mi(long long int a, long long int m) {
  long long int x, y;
  eea(a, m, x, y);
  x = (x % m + m) % m;
  return x;
}
void Sieve(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++) {
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    long long int n, p;
    cin >> n >> p;
    int x = 0, y = 0;
    for (long long int i = 0;; i++) {
      if (n * n > (x * x + y * y) * (p * p)) {
        if (i % 2 == 0)
          x++;
        else
          y++;
      } else
        break;
    }
    if ((x + y - 1) % 2 == 0)
      cout << "Utkarsh\n";
    else
      cout << "Ashish\n";
  }
}
