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 <math.h>
#include <stdio.h>

#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>

#define int ll
#define PI acos(-1.0)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> pii;
typedef vector<pii> vpii;

ll t;
ll d, k;

int shem(ll d) {
// d =
    ll tmp;
    ll n = d / 2;
    ll sq = (int)(sqrt(n) - 1);
    if (sq < 0) sq = 0;
    for (n = d / 2; ; n++) {
        while(sq * sq <= n) sq++;
        sq--;
        tmp = n + sq * sq;
        if (tmp == d) return 2;
        if (tmp > d) return 1;
    }
    return false;
}

signed main() {
    cout.precision(15);
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin>>t;
    while (t--) {
        cin>>d>>k;
        d = d * d / k / k;
        int pas = shem(d);
        cout << (pas == 2 ? "Utkarsh" : "Ashish") << endl;
    }
    return 0;
}
