Codeforce-- 思维题

2-11 Eat the chip

思路：题意可知，a从下往上走，b从上往下走，我们可以思考一个问题，在某一个回合，如果a和b在同一列中（a是Alice，b是Bob），一方怎么走，另外一方就跟着这一方走，所以同一列的时候是该回合后手的必胜态，先手的必败态。于是可以递推：如果在走到边界的时候有机会仍然保持b在a的上方，那么有机会分出胜负，因为一方不可以走出边界，所以ya-yb一定会发生变化，直到xa-xb = 0

import sys
#file I/O
#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')

#faster I/O
input = sys.stdin.readline
sys.setrecursionlimit(30)

def print(ans='',end = '\n'):
    sys.stdout.write(str(ans) + end)

def solve() -> None:
    n,m,xa,ya,xb,yb = map(int,input().split())
    flag = bool((xb - xa) % 2)
    if not flag:
        winner = "Bob"
        if xa >= xb:
            win = False
        elif ya == yb:
            win = True
        else:
            if yb > ya:
                n_turn = yb - 1
            else:
                n_turn = m - yb
            win = xb - 2 * n_turn >= xa
    else:
        winner = "Alice"
        xa += 1
        ya += 0 if yb - ya == 0 else 1 if yb - ya > 0 else -1
        if xa > xb:
            win = False
        elif ya == yb:
            win = True
        else:
            if ya < yb :
                n_turn = m - ya
            else:
                n_turn = ya - 1
            win = xb - 2 * n_turn >= xa
    print(winner if win else "Draw")



T = int(input())
for _ in range(T):
    solve()