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>
typedef long long ll;
using namespace std;
const int maxn = 2e5+10;
const int mod = 1e9+7;
typedef pair<int,int> pii;
void solve()
{
    ll d,k;
    cin >> d >> k;
    d*=d;k*=k;d/=k;
    ll now = 1;
    d--;
    if(d<=0)cout <<"Ashish" << endl;     
    while(d>0){
        now++;
        ll pre = now-1;
        ll o = now*now - pre*pre;
        d-=o;
        if(d<=0)cout << "Utkarsh" << endl;
        else {
            d-=o;
            if(d<=0)cout <<"Ashish" << endl;  
        } 
    }
    return ;
}
int main(){
    int T;
    for(cin >> T;T;T--){
        solve();
    }
    // system("pause");
}