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 long long inf = 1e9;
const long long inf64 = 1e18;
const long long MOD = inf + 7;
long long d, k;
map<pair<long long, pair<long long, long long> >, bool> mp;
bool ok(long long x, long long y, long long turn) {
  if (x * k * x * k + y * k * y * k > d * d) {
    return 0;
  }
  if (mp.find({x, {y, turn}}) != mp.end()) {
    return mp[{x, {y, turn}}];
  }
  long long res = 0;
  if (turn == 0) {
    if (ok(x + 1, y, 1) == 0 || ok(x, y + 1, 1) == 0) res = 1;
  } else {
    if (ok(x + 1, y, 0) == 0 || ok(x, y + 1, 0) == 0) res = 1;
  }
  mp[{x, {y, turn}}] = res;
  return res;
}
void solve() {
  cin >> d >> k;
  mp.clear();
  cout << (ok(0, 0, 0) == 0 ? "Ashish" : "Utkarsh") << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
}
