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>
t=int(input())
for _ in range(t):
    d,k=[int(i) for i in input().split()]
    a=0;b=0
    cd=0
    while a*a+b*b<=d*d:
        if (a+k+k)*(a+2*k)+(b+k)*(b+k)>d*d:
           a+=k
           if cd==0:cd=1
           else:cd=0
        if (b+2*k)*(b+2*k)+(a+k)*(a+k)>d*d:
           b+=k
           if cd==0:cd=1
           else:cd=0
        if a<b:
           a+=k
           if cd==0:cd=1
           else:cd=0
        else:
           b+=k
           if cd==0:cd=1
           else:cd=0
    if cd==1:print("Utkarsh")
    else:print("Ashish")