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;
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p);
template <class T>
ostream &operator<<(ostream &os, vector<T> &v);
template <class T>
ostream &operator<<(ostream &os, set<T> &v);
inline void optimizeIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
const int nmax = 2e5 + 7;
void solveTC() {
  int d, k;
  cin >> d >> k;
  if (k == 1) {
    int p = (d / k) % 2;
    if (p == 1)
      cout << "Ashish"
           << "\n";
    else {
      cout << "Utkarsh"
           << "\n";
    }
    return;
  }
  int mv = (d * d) / (k * k);
  int p = mv % 2;
  if (p == 0)
    cout << "Ashish"
         << "\n";
  else {
    cout << "Utkarsh"
         << "\n";
  }
}
int32_t main() {
  optimizeIO();
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solveTC();
  }
  return 0;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << "{" << p.first << ", " << p.second << "} ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "[ ";
  for (T i : v) {
    os << i << " ";
  }
  os << " ]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, set<T> &v) {
  os << "[ ";
  for (T i : v) {
    os << i << " ";
  }
  os << " ]";
  return os;
}
