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>
from __future__ import division,print_function
from heapq import*
import sys
from math import*

le = sys.__stdin__.read().split("\n")[::-1]
af = []
for i in range(int(le.pop())):
    d,k=list(map(int,le.pop().split()))
    a=floor(d/(k*2**0.5))
    #print(a,d,k)
    if (a*k)**2+(k*(a+1))**2>d**2:
        af.append('Utkarsh')
    else:
        while (a*k)**2+(k*(a+2))**2>d**2 and a>0:
            a-=1
        if 2*k**2>d**2:
            af.append('Ashish')
        elif ((a+1)*k)**2+(k*(a+2))**2>d**2:
            #print("couc")
            af.append('Utkarsh')
        else:
            af.append("Ashish")
print("\n".join(map(str,af)))
