import math
#需要用一个math里面的ceil函数

def fastsort(lister):
    lengh = len(lister)
    for i in range(1, lengh):
        pre = lister[i - 1][2]
        while i >= 1 and lister[i][2] < pre:
            lister[i - 1][2] = lister[i][2]
            lister[i][2] = pre
            i -= 1
            pre = lister[i - 1][2]
    return lister
#简单的排序算法
def timecount(ming,demon):
    hp = ming[0]
    ark = ming[1]
    hp_demon = demon[0]
    ark_demon = demon[1]
    count_ming = math.ceil(hp_demon / ark)
    count_demon = math.ceil(hp / ark_demon)
    if count_ming < count_demon:
        k = 1
    else:
        k = 0

    hp -= count_ming * ark_demon
    return [k, count_ming, hp]
# k 是帮助判断的值，k = 0 说明一鸣师姐打不过怪物，hp则是返回战斗后剩余hp，count———ming是一个帮助下一步判断的变量，它是一鸣师姐击败怪物需要的时间

ming = [int(i)for i in str(input()).split()]
n = int(input())
lister = []
resumetime = ming[2]
hp = ming[0]
origion_hp = ming[0]
ark = ming[1]
judge = 1
#judge 也是一个帮助判断的变量
for i in range(n):
    demon = [int(i)for i in str(input()).split()]
    lister.append(demon)
#处理一下数据，按照时间排序一下
lister = fastsort(lister)
alter = 0
for i in lister:
    alter += 1
    data_return = timecount(ming, i)
    k = data_return[0]
    sumtime = data_return[1]
    ming[0] = data_return[2]
    time = i[2]
    if alter <= len(lister) - 1:
        curtime = lister[alter - 1][2] + sumtime
        nexttime = lister[alter][2]
    else:
        curtime = lister[alter - 1][2] + sumtime
        nexttime = curtime

    if k == 0:
        print('My Milk!')
        judge = 0
        break
    elif curtime > nexttime:
        print('My Milk!')
        judge = 0
        break
    elif nexttime - curtime >= resumetime:
        hp = ming[0]
        hp += ming[3]
        ming[0] = min(hp, origion_hp)
# 一鸣失败有两种情况，一是打不过怪物成眠了，函数返回k是0，另外一种就是同时多个怪物，就是利用了curtime和nexttime两个变量判断，前者是打败这只怪物的时候，后者是下一个怪物出现的时候，如果curtime 》 nexttime说明会同时出现多个怪物
if judge == 1:
    print('Mission Complete.')