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;
using ll = long long;
using ld = long double;
using pr = pair<int, int>;
using prll = pair<long, long>;
using V = vector<int>;
using Vll = vector<long long>;
template <typename T, typename TT>
pr operator+(pair<T, TT> a, pair<T, TT> b) {
  return {a.first + b.first, a.second + b.second};
}
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (auto itr : t) s << itr << " ";
  return s;
}
long long cal(int x) { return x * 1LL * x; }
int bin_search(int d, int k) {
  int beg = 0;
  int last = d;
  long long cmp = d * 1LL * d;
  long long clp = k * 1LL * k;
  while (beg < last) {
    int mid = beg + (last - beg + 1) / 2;
    if (cal(mid) * clp <= cmp)
      beg = mid;
    else
      last = mid - 1;
  }
  return beg;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    int d, k;
    cin >> d >> k;
    int ans = bin_search(d, k);
    long long temp = ans * 1LL * (ans + 1) * k * k;
    if (temp < d * 1LL * d)
      cout << "Ashish\n";
    else
      cout << "Utkarsh\n";
  }
  return 0;
}
