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>
import sys
# from math import sqrt
# import re
# from heapq import *
# from collections import defaultdict as dd
# from collections import OrderedDict as odict
# from collections import Counter as cc
# from collections import deque
# sys.setrecursionlimit(3*(10**5)+1)#thsis is must
# mod = 10**9+7; md = 998244353
# m = 2**32
input = lambda: sys.stdin.readline().strip()
inp = lambda: list(map(int,sys.stdin.readline().strip().split()))
# def C(n,r,mod):
#   if r>n:
#       return 0
#   num = den = 1
#   for i in range(r):
#       num = (num*(n-i))%mod
#       den = (den*(i+1))%mod
#   return (num*pow(den,mod-2,mod))%mod
# def pfcs(M = 100000):
#   pfc = [0]*(M+1)
#   for i in range(2,M+1):
#       if pfc[i]==0:
#           for j in range(i,M+1,i):
#               pfc[j]+=1
#   return pfc
#______________________________________________________
t,=inp()
while(t>0):
    # int d, k;
    d,k = inp()
    x = 0; y = 0;
    while(True):
        if(x <= y and (x + k) * (x + k) + y * y <= d * d):
            x += k;
        elif(x > y and (y + k) * (y + k) + x * x <= d * d):
            y += k;
        else:
            break;
    if(x == y):
        print("Utkarsh")
    else:
        print("Ashish")
    t-=1