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 <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <fstream>
#include <bitset>
#include <cassert>
#include <functional>
#include <cstring>
#include <ctime>
#include <cmath>
#include <set>
#include <map>
#define sec second
#define fir first
#include <algorithm>
#define pb push_back
#define mp make_pair
#define REP(i , n) for (int i = 0 ; i < (n) ; i++)
#define BACK(i , n) for (int i = (n) - 1 ; i >= 0 ; i--)
using namespace std;
using vi = vector<int>;
using vii = vector<vi>;
using pii = pair<int , int>;
using ll = long long;
using ld = long double;
const int MAX = 1e9 , MIN = - MAX , MOD = 1e9 + 7;
const ll MAX_LL = 1e18 , MIN_LL = -MAX_LL;
int main(){
    int T;
    cin >> T;
    while(T--){
        int d , k , maks = 0;
        cin >> d >> k;
        for (int i = 0 ; i <= d ; i += k){
            maks = max(maks , i / k + (int)sqrtl((ll)d * d - (ll)i * i) / k);
        }
        cout << (maks & 1 ? "Ashish" : "Utkarsh") << '\n';
    }
}
