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;
const int N = 100001;
const int M = 20;
const int MOD = 1000000007;
const int inf = 1234567890;
const long long INF = 1122334455667788990;
long long solve(long long k, long long x) {
  long long lo = 0, hi = x;
  while (lo < hi) {
    long long m = (lo + hi) / 2;
    if ((((m * k)) * ((m * k))) <= x)
      lo = m + 1;
    else
      hi = m - 1;
  }
  return lo - 1;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int tc = 1, cs = 0;
  cin >> tc;
  while (tc--) {
    long long d, k, i, mx = 0;
    cin >> d >> k;
    for (i = 1; i <= (d / k); i++) {
      long long t = ((d) * (d)) - (((i * k)) * ((i * k)));
      long long x = solve(k, t);
      mx = max(mx, i + x);
    }
    if (mx % 2LL)
      cout << "Ashish" << '\n';
    else
      cout << "Utkarsh" << '\n';
  }
}
