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;
bool findans(map<int64_t, bool> &dist, map<pair<int, int>, bool> &visited,
             int64_t d, int64_t x, int64_t y, int stride) {
  if (x * x + y * y > d * d) return false;
  if (visited[{x, y}] == true || visited[{y, x}] == true) {
    return dist[x * x + y * y];
  }
  dist[x * x + y * y] = false;
  dist[x * x + y * y] = !(findans(dist, visited, d, x + stride, y, stride) &&
                          findans(dist, visited, d, x, y + stride, stride));
  visited[{x, y}] = true;
  visited[{y, x}] = true;
  return dist[x * x + y * y];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int64_t d, k;
    cin >> d >> k;
    map<int64_t, bool> dist;
    map<pair<int, int>, bool> visited;
    bool ans = findans(dist, visited, d, 0, 0, k);
    if (ans) {
      cout << "Utkarsh\n";
    } else {
      cout << "Ashish\n";
    }
  }
  return 0;
}
