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;
const int maxn = 1e6 + 5;
int k = 2, d = 5;
bool dfs(int y, int x) {
  if (x * x + y * y > d * d) return true;
  if (dfs(y + k, x) && dfs(y, x + k)) return false;
  return true;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> d >> k;
    long long x = 0;
    while ((x + k) * (x + k) * 2 <= d * d) x += k;
    long long y = x + k;
    if (x * x + y * y <= d * d)
      cout << "Ashish" << endl;
    else
      cout << "Utkarsh" << endl;
  }
  return 0;
}
