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>
# Design_by_JOKER

from math import *
#from cmath import *
from itertools import *
from decimal import *  # su dung voi so thuc
from fractions import *  # su dung voi phan so
from sys import *
#from numpy import *

'''getcontext().prec = x # lay x-1 chu so sau giay phay (thuoc decimal)
Decimal('12.3') la 12.3 nhung Decimal(12.3) la 12.30000000012
Fraction(a) # tra ra phan so bang a (Fraction('1.23') la 123/100 Fraction(1.23) la so khac (thuoc Fraction)
a = complex(c, d) a = c + d(i) (c = a.real, d = a.imag)
a.capitalize() bien ki tu dau cua a(string) thanh chu hoa, a.lower() bien a thanh chu thuong, tuong tu voi a.upper()
a.swapcase() doi nguoc hoa thuong, a.title() bien chu hoa sau dau cach, a.replace('a', 'b', slg)
chr(i) ki tu ma i ord(c) ma ki tu c
a.join['a', 'b', 'c'] = 'a'a'b'a'c, a.strip('a') bo dau va cuoi ki tu 'a'(rstrip, lstrip)
a.split('a', slg = -1) cat theo ki tu 'a' slg lan(rsplit(), lsplit()), a.count('aa', dau = 0, cuoi= len(a)) dem slg
a.startswith('a', dau = 0, cuoi = len(a)) co bat dau bang 'a' ko(tuong tu endswith())
a.index("aa") vi tri dau tien xuat hien (rfind())
input = open(".inp", mode='r')  a = input.readline()
out = open(".out", mode='w') a.index(val) '''

for _ in range(int(input())):
	d, k = map(int, input().split())
	z = int(sqrt((d*d)//(k*k*2)))
	print(['Ashish', 'Utkarsh'][k*k*(z*z+z*z+2*z+1) > d*d])