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 long long N = 1111;
long long d, k;
long long solve() {
  cin >> d >> k;
  long long maxi = d * d / (k * k);
  set<pair<long long, long long> > st;
  for (long long al = (0); al <= (1000000); al++) {
    long long need = maxi - al * al;
    if (need < 0) {
      break;
    }
    long long x = sqrt(need);
    st.insert({al, x});
    st.insert({x, al});
  }
  long long x = -1;
  for (auto p : st) {
    if (p.first == p.second) {
      x = p.first;
    }
  }
  if (x != -1) {
    for (auto p : st) {
      if (p.first != p.second && (p.first == x - 1 && p.second == x)) {
        x = -1;
      }
    }
  }
  if (x == -1) {
    cout << "Ashish"
         << "\n";
  } else {
    cout << "Utkarsh"
         << "\n";
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
