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>
template <typename T1, typename T2>
inline bool chkmin(T1& x, const T2& y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool chkmax(T1& x, const T2& y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
using namespace std;
const int maxn = 2e5 + 20;
ostream& operator<<(ostream& out, vector<int>& v) {
  for (auto& x : v) out << x << ' ';
  return out;
}
ostream& operator<<(ostream& out, pair<int, int>& v) {
  out << v.first << ", " << v.second;
  return out;
}
istream& operator>>(istream& in, pair<int, int>& a) {
  in >> a.first >> a.second;
  return in;
}
const long long inf = (long long)2e9;
const long double pi = asin(1) * 2;
const long double eps = 1e-8;
const long long mod = (long long)1e9 + 7;
const long long ns = 97;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  int d, k;
  cin >> d >> k;
  vector<pair<int, int> > one, two;
  for (int b = 0; b < d + 1; ++b) {
    int q = b * k;
    if (q <= d) {
      int a = sqrt(d * d - q * q) / k;
      int t1 = a * k, t2 = (b + 1) * k;
      if (t1 * t1 + t2 * t2 > d * d) {
        if ((a + b) % 2 == 0)
          two.push_back({a, b});
        else
          one.push_back({a, b});
      }
    }
  }
  if ((int)one.size() > (int)two.size())
    cout << "Ashish";
  else
    cout << "Utkarsh";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int q = 1;
  cin >> q;
  while (q--) solve();
}
