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<map>
#include<unordered_map>
#include<queue>
#include<stack>
#include<vector>
#include <cmath>
#include<math.h>
#define ll long long int
#define pb push_back
#define mod 1000000007
#define ip(ar,m,n) for(ll i=0;i<m;i++){for(ll j=0;j<n;j++){cin>>ar[i][j];}}
#define mk(ar,m,n) for(ll i=0;i<m;i++){ar[i]=new ll[n];}
using namespace std;

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
double dis(ll x,ll y){
    return sqrt(x*x+y*y);
}

void solve(ll r,ll z,ll k){
    double d=dis(z,z+k);
    if(d<=r){
        cout<<"Ashish"<<endl;
    }
    else{
        cout<<"Utkarsh"<<endl;
    }
}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll t;
    cin>>t;
    while(t--){
        double rt_2=sqrt(2);
        double r,k,z;
        cin>>r>>k;
        z=r/rt_2;
        int tmp=(int)z,tmp2=(int)k;
        z=tmp-tmp%tmp2;
        solve(r,z,k);
    }
    return 0;
}
