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>
//#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
//using namespace __gnu_pbds;


typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;


#define pb push_back
#define ff first
#define ss second
#define INF int(2e9 + 9)
#define BIG_INF ll(2e18)
#define SIZE 200010
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define len(v) ll((v).size())
#define BACK(v) ((v).back())
#define sqr(x) (x) * (x)
#define cube(x) (x) * (x) * (x)
#define filein freopen("input.txt", "r", stdin)
#define fileout freopen("output.txt", "w", stdout);

//#pragma GCC target ("avx2")
//#pragma GCC optimization ("O3")
//#pragma GCC optimization ("unroll-loops")


ll binpow(ll a, ll b, ll MOD = BIG_INF)
{
    ll res = 1;
    a %= MOD;
    while(b)
    {
        if(b & 1)
            res *= a;
        a *= a;
        b >>= 1;
        a %= MOD;
        res %= MOD;
    }

    return res % MOD;
}
template<typename T> void umax(T& a, T b){a = max(a, b);}
template<typename T> void umin(T& a, T b){a = min(a, b);}


template<typename T>
ostream& operator<<(ostream& s, vector<T>& v)
{
    for (auto el : v)
    {
        s << el << " ";
    }
    return s;
}


const ll mod = 998244353;

int T = 1;
ll n, m, k, d;
string s;


void solve()
{

    cin >> d >> k;

    ll a = 1;

    while(sqr(a * k) * 2 <= sqr(d)) a++;

    a--;

    if(sqr(a * k) + sqr((a + 1) * k) > sqr(d))
    {
        cout << "Utkarsh\n";
    }else cout << "Ashish\n";


}



int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

//    filein;fileout;

    cin >> T;

    while(T--)
    {
        solve();
    }
    return 0;
}
