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 unsigned mod = 998244353;
const unsigned _mod = 1e9 + 7;
const long long infi = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % mod;
    y >>= 1ll;
    x = x * x % mod;
  }
  return ret;
}
long long qpow(long long x, long long y, long long m) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % m;
    y >>= 1ll;
    x = x * x % m;
  }
  return ret;
}
long long _gcd(long long x, long long y) { return y ? _gcd(y, x % y) : x; }
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long l, r, g;
long double p;
int main() {
  p = sqrt(2);
  int T;
  cin >> T;
  while (T--) {
    cin >> l >> r;
    g = trunc(l * 2 / p);
    g /= r;
    if (g & 1ll)
      puts("Ashish");
    else
      puts("Utkarsh");
  }
  return 0;
}
