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;
void printv(vector<long long int> v) {
  long long int sz = v.size();
  for (long long int i = 0; i < sz; i++) {
    if (i == sz - 1)
      cout << v[i] << "\n";
    else
      cout << v[i] << " ";
  }
}
void input(long long int a[], long long int sz) {
  for (long long int i = 0; i < sz; i++) cin >> a[i];
}
void print(long long int a[], long long int sz) {
  for (long long int i = 0; i < sz; i++) {
    if (i == sz - 1)
      cout << a[i] << "\n";
    else
      cout << a[i] << " ";
  }
}
long long int dx1[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long int dy1[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long int dx2[] = {1, 0, -1, 0};
long long int dy2[] = {0, 1, 0, -1};
long long int mul(long long int a, long long int b,
                  long long int p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long int add(long long int a, long long int b,
                  long long int p = 1000000007) {
  return (a % p + b % p) % p;
}
long long int power(long long int x, long long int y,
                    long long int p = 1000000007) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p = 1000000007) {
  return power(n, p - 2, p);
}
long long int ncrMod(long long int n, long long int r,
                     long long int p = 1000000007) {
  if (r == 0) return 1;
  long long int fac[n + 1];
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int convert(string num) {
  long long int sum = 0;
  for (int i = 0; i < num.size(); i++) {
    sum = sum * 10;
    sum += (num[i] - '0');
  }
  return sum;
}
bool ispeed2ime(long long int n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
bool isPowerof2(long long int n) {
  while (n > 1) {
    if (n % 2 != 0) {
      return true;
    }
    n /= 2;
  }
  return false;
}
void solve() {
  long long int d, k;
  cin >> d >> k;
  long long int temp = d * d;
  long long int x = 0, y = 0, cnt = 0, flag = 0;
  while (1) {
    long long int mini = min(x, y);
    mini += k;
    long long int maxi = max(x, y);
    if (maxi * maxi + mini * mini > temp) {
      break;
    }
    x = maxi;
    y = mini;
    flag ^= 1;
  }
  cout << (flag == 1 ? "Ashis" : "Utkarsh") << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
