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;
#define ll long long 
#define pii pair<int,int>
#define fi first 
#define se second 
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define pb push_back
#define faster ios_base::sync_with_stdio(0);cin.tie(0);
#define mem(a,v) memset(a,v,sizeof(a))
const ll INF=1e18;
const int mod=1e9+7;
const int iinf=1e9+10;
const int xd[4]={1,0,-1,0},yd[4]={0,1,0,-1};
ll d,k;
void solve(){
    cin>>d>>k;
    ll x=(d*d)/(k*k);
    ll s=0,i=1,j=1;
    while(1){
        s+=i*i-(i-1)*(i-1);
        if(s>x){
            cout<<"Utkarsh\n";return;
        }
        //cout<<"ss="<<s<<"\n";
        //if(s+(i+1)*(i+1)-i*i<x&&s+(i+1)*(i+1)-i*i+j*j-(j-1)*(j-1)>x){
           // cout<<"Ashishe\n";return;
        //}
        s+=j*j-(j-1)*(j-1);
        if(s>x){
            cout<<"Ashish\n";return;
        }
        //cout<<"s="<<s<<"\n";
        i++;j++;
    }
}
int main()
{
    faster
    int t;
    cin>>t;
    while(t--)
        solve();
}