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;
inline bool valid(int x, int n) { return 0 <= x && x < n; }
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
template <typename T>
inline T pop(queue<T>& q) {
  T front = q.front();
  q.pop();
  return front;
}
template <typename T>
inline T gcd(T a, T b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
template <typename T>
tuple<T, T, T> xgcd(T a, T b) {
  if (b == 0) return {1, 0, a};
  T x, y, g;
  tie(x, y, g) = xgcd(b, a % b);
  return {y, x - (a / b) * y, g};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int d, k;
    cin >> d >> k;
    long long int r = d;
    long long int lo = 0;
    long long int hi = d;
    while (lo < hi) {
      long long int mid = (lo + hi) / 2;
      if (mid * mid * 2 > d * d) {
        hi = mid;
      } else {
        r = max(r, mid * 2);
        lo = mid + 1;
      }
    }
    if (k > r / 2) {
      cout << "Ashish\n";
      continue;
    }
    if ((r / k) % 2)
      cout << "Ashish\n";
    else
      cout << "Utkarsh\n";
  }
}
