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;
long long t;
long long d, k;
struct node {
  long long x, y;
  bool operator<(const node &t) const {
    if (y != t.y) return y < t.y;
    return x < t.x;
  }
};
vector<node> vv, v;
bool check_node(node a) {
  return a.x * a.x * k * k + a.y * a.y * k * k <= d * d;
}
bool check(node a) {
  if (check_node({a.x, a.y + 1})) return 0;
  if (check_node({a.x + 1, a.y})) return 0;
  return 1;
}
signed main() {
  cin >> t;
  while (t--) {
    cin >> d >> k;
    vv.clear();
    v.clear();
    for (long long i = d / k, j = 0; j <= d / k; j++) {
      while (i * i * k * k + j * j * k * k > d * d) i--;
      vv.push_back({i, j});
    }
    for (long long i = 0, j = d / k; i <= d / k; i++) {
      while (i * i * k * k + j * j * k * k > d * d) j--;
      vv.push_back({i, j});
    }
    for (auto now : vv) {
      if (check(now)) {
        v.push_back(now);
      }
    }
    sort(v.begin(), v.end());
    node pre = {-1, -1};
    long long c = -1, len, max0 = -1, max1 = -1;
    for (auto now : v) {
      if (now.x == pre.x && now.y == pre.y) continue;
      long long v = (now.x + now.y) % 2;
      if (v != c) {
        if (c == 0) max0 = max(max0, len);
        if (c == 1) max1 = max(max1, len);
        c = v;
        len = 1;
      } else {
        len++;
      }
      pre = now;
    }
    if (c == 0) max0 = max(max0, len);
    if (c == 1) max1 = max(max1, len);
    if (max0 > max1) {
      puts("Utkarsh");
    } else {
      puts("Ashish");
    }
  }
  return 0;
}
