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;    
typedef long long ll;
typedef long double ld;
#define lc (v<<1)    
#define rc (lc|1)
#define tm ((tl+tr)>>1)
#define pll pair<ll,ll>
#define fr first
#define sc second
#define low lower_bound
#define upp upper_bound
const ll mxn = 1e5+10;
const ll inf = 1e9+7;
const ll mod = 1e9+7;

ll gcd(ll a, ll b){
    return !b?a:gcd(b,a%b); 
}
     
ll lcm(ll a, ll b){return (a / gcd(a, b)) * b;}

ll t;
vector<ll>v;

ll f(ld d, ld k){
    ll ans=0;
    while(d > sqrt(2*k*k))d-=sqrt(2*k*k) , ans++;
    return ans;
}

int main(){

    cin >> t;

    while(t--){
        ld d, k;
        cin >> d >> k;
        if(k == d)cout << "Ashish" << endl;
        else{
            ll ans = f(d,k);
            ld ans2 = (ans*ans*k*k)+((ans+1)*(ans+1)*k*k);
            if(ans2 <= d*d)cout << "Ashish" << endl;
            else cout << "Utkarsh" << endl;
        }
    }

    return 0;
}