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
from sys import stdin, stdout
# from fractions import gcd
# from math import *
# from operator import mul
# from functools import reduce
# from copy import copy
from collections import deque, defaultdict, Counter

rstr = lambda: stdin.readline().strip()
rstrs = lambda: [str(x) for x in stdin.readline().split()]
rint = lambda: int(stdin.readline())
rints = lambda: [int(x) for x in stdin.readline().split()]
rstr_2d = lambda n: [rstr() for _ in range(n)]
rint_2d = lambda n: [rint() for _ in range(n)]
rints_2d = lambda n: [rints() for _ in range(n)]
pr = lambda args, sep: stdout.write(sep.join(map(str, args)) + '\n')
out = []

for _ in range(int(input())):
    d, k = rints()
    cur, ans = 0, 0
    a = [i for i in range(k, d, k)]
    d, be, en = pow(d, 2), 1, len(a)

    while be <= en:
        md = (be + en) >> 1
        val = pow(a[md - 1], 2) << 1
        if val <= d:
            cur = a[md - 1]
            be = md + 1
        else:
            en = md - 1

    if pow(cur + k, 2) + pow(cur, 2) <= d:
        ans = 1

    out.append(['Utkarsh', 'Ashish'][ans])

pr(out, '\n')
