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>
//FIRST THINK THEN CODE.

#include <bits/stdc++.h>


using namespace std;

typedef long long ll;

#define rep(i,a,b) for(ll i=a;i<b;++i)
#define rrep(i,a,b) for(ll i=a;i>b;--i)
#define FOR(i,n)  for(ll i=0;i<n;i++)
#define vi vector<int>
#define vl vector<ll>
#define ld long double
#define vld vector<ld>
#define vvi vector<vector<int>>
#define vvl vector<vector<long long>>
#define vvld vector<vector<ld>>
#define pii pair<int,int>
#define pll pair<long,long>
#define vpii vector<pii>
#define vpll vector<pll>
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define d1(x) cout<<(x)<<endl
#define d2(x,y) cout<<(x)<<" "<<(y)<<endl
#define d3(x,y,z) cout<<(x)<<" "<<(y)<<" "<<(z)<<endl
#define d4(a,b,c,d) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<endl
#define d6(a,b,c,d,e,f) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<" "<<(e)<<" "<<(f)<<endl
#define PI 3.1415926535897932384626433832795
#define fix(f,n) fixed<<setprecision(n)<<f
#define all(x) x.begin(),x.end()
#define rev(p) reverse(p.begin(),p.end());
#define endl "\n"
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define popcount(x) __builtin_popcountll(x)
#define sz(x) ((ll)x.size())
const ll M = 1000000007;
const ll MM = 998244353;
ll begtime = clock();
#define end_routine() cerr << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//#define trace(...)
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1 << " | "; __f(comma + 1, args...);
}

template<typename T, typename F>
void chmax( T &a, F b) {
  if (b > a)a = b;
}

template<typename T, typename F>
void chmin( T &a, F b) {
  if (b < a)a = b;
}



/* if you want to multiply 2 64 bit numbers mod c.
 i.e a*b%c.
 this function will help  in preventing "overflow".
*/

ll mulmod(ll a, ll b, ll c) {

  ll ans = 0;
  ll y = a % c;
  while (b) {
    if (b & 1) {
      (ans += y) %= c;
    }
    y = y * 2 % c;
    b >>= 1;
  }
  return ans;

}



ll powM(ll a, ll b, ll m)
{
  if (b < 0)return 0;
  if (a <= 0)return 0;
  a %= m;

  ll ans = 1LL;
  while (b)
  {
    if (b & 1)ans = ans * a % m;
    //ans = mulmod(ans, a, m);
    a = a * a % m;
    //a = mulmod(a, a, m);
    b >>= 1;
  }

  return ans;
}

ll powMbig(ll a, ll b, ll m)
{
  if (a <= 0)return 0;
  a %= m;

  ll ans = 1LL;
  while (b)
  {
    if (b & 1)//ans = ans * a % m;
      ans = mulmod(ans, a, m);
    //a = a * a % m;
    a = mulmod(a, a, m);
    b >>= 1;
  }

  return ans;
}




ll poww(ll a, ll b)
{

  ll ans = 1;
  while (b)
  {
    if (b & 1)ans = ans * a;
    a = a * a;
    b >>= 1;
  }

  return ans;

}


string tostring(ll x) {
  stringstream sss;
  sss << x;
  string ans = sss.str();
  return ans;
}



const ll N = 2e5 + 5;




vl a;
vl sg;


void build(ll x, ll lx, ll rx) {
  if (lx == rx) {
    sg[x] = a[lx];

    return;
  }
  ll mx = (lx + rx) / 2;
  build(2 * x, lx, mx);
  build(2 * x + 1, mx + 1, rx);
//  combine(x);
  sg[x] = sg[2 * x] + sg[2 * x + 1];
}

void update(ll x, ll lx, ll rx, ll idx, ll v) {
  if (lx == rx) {
    sg[x] = v;
    // sg[x].pr = sg[x].sf = sg[x].res = max(0ll, v);
    return;
  }
  ll mx = (lx + rx) / 2;
  if (idx <= mx) {
    update(2 * x, lx, mx, idx, v);
  } else {
    update(2 * x + 1, mx + 1, rx, idx, v);
  }
  sg[x] = sg[2 * x] + sg[2 * x + 1];
// combine(x);
}


// find idx st sum[0...idx]==k.

ll calc(ll x, ll lx, ll rx, ll l, ll r) {


  if (lx > r || rx < l)return 0;

  if (lx >= l && rx <= r)return sg[x];

  ll mx = (lx + rx) / 2;


  return calc(2 * x + 1, mx + 1, rx, l, r) + calc(2 * x, lx, mx, l, r);


}


int main() {

  IOS;

#ifndef ONLINE_JUDGE

  freopen("input1.txt", "r", stdin);
  freopen("output1.txt", "w", stdout);

#endif

  ll t; cin >> t;

  while (t--) {

    ll d, k;
    cin >> d >> k;

    ll r1 = 0;
    ll a1 = 0;
    while (2 * r1 * r1 * k * k <= d * d) {
      a1 = 2 * r1 * r1 * k * k;
      r1++;
    }

    ll r2 = 0;
    ll a2 = 0;
    while ((2 * r2 * r2 + 4 * r2 + 4)*k * k <= d * d) {
      a2 = (2 * r2 * r2 + 4 * r2 + 4) * k * k;
      r2++;
    }


    ll r3 = 0;
    ll a3 = 0;
    while ((2 * r3 * r3 + 2 * r3 + 1)*k * k <= d * d) {
      a3 = (2 * r3 * r3 + 2 * r3 + 1) * k * k;
      r3++;
    }

    if (a3 >= max(a1, a2)) {
      d1("Ashish");
    } else d1("Utkarsh");


  }

  return 0;
}