//贪心算法解决加油站选择问题

//# include<iostream>

# include<stdio.h>

using namespace std;

# include<algorithm>

struct Node

{

    float p, d;

};

bool cmp(Node a, Node b)

{

    return a.d < b.d;

}

int main()

{

    Node node[];

    float Cmax, D, Davg, distance, price, Ccur, Pcur, j;//double 会出问题，蛋疼

    int N, Ncur, i, k, flag;

    //while (cin >> Cmax >> D >> Davg >> N)

    while (scanf_s("%f%f%f%d", &Cmax, &D, &Davg, &N) != EOF)

    {

        for (i = ; i <= N; i++)

        {

            //cin >> node[i].p >> node[i].d;

            scanf_s("%f%f", &node[i].p, &node[i].d);

        }

        //sort station by distance

        sort(node + , node +  + N, cmp);

        if (N ==  || node[].d > 0.00001 || node[].d < -0.00001)//第一个站点不在距离0处

        {

            printf("The maximum travel distance = 0.00\n");

        }

        else//greedy now

        {

            price = ;//当前花费

            distance = ;//当前距离

            Ncur = ;//当前所在加油站

            Ccur = ;//当前汽油数量

            Pcur = node[].p;//当前加油站的油价

            while (Ncur <= N)

            {

                //1.using 【remain】 gas find 【nearest】 【cheaper】 station

                flag = -;

                for (i = Ncur + ; i <= N&&node[i].d <= distance + Ccur*Davg; i++)

                {

                    if (node[i].p < Pcur)

                    {

                        flag = i;

                        break;

                    }

                }

                if (flag != -)//find a station,and get there

                {

                    //price 未变

                    Ccur -= (node[flag].d - distance) / Davg;//当前汽油数量===注意与下面表达式的计算顺序

                    distance = node[flag].d;//当前距离

                    Ncur = flag;//当前所在加油站

                    Pcur = node[flag].p;//当前加油站的油价

                    continue;

                }

                //else 1: 2.using 【Cmax】 find 【nearest】 【cheaper】 station

                flag = -;

                for (i = Ncur + ; i <= N&&node[i].d <= distance + Cmax*Davg; i++)

                {

                    if (node[i].p < Pcur)

                    {

                        flag = i;

                        break;

                    }

                }

                if (flag != -)//find a station,and get there

                {

                    price += ((node[flag].d - distance) / Davg - Ccur)*node[Ncur].p;//当前花费

                    distance = node[flag].d;//当前距离

                    Ncur = flag;//当前所在加油站

                    Pcur = node[flag].p;//当前加油站的油价

                    Ccur = ;//当前汽油数量：到K后的汽油为0

                    continue;

                }

                //else 2: 3.get the Cmax and go as far as possible

                flag = -;

                for (i = Ncur + ; i <= N&&node[i].d <= distance + Cmax*Davg; i++)

                {

                    flag = i;

                }

                if (flag != -)//can get some one station,and get there

                {

                    price += (Cmax - Ccur)*node[Ncur].p;//当前花费

                    distance = node[flag].d;//当前距离

                    Ncur = flag;//当前所在加油站

                    Pcur = node[flag].p;//当前加油站的油价

                    Ccur = Cmax - (node[flag].d - distance) / Davg;//当前汽油数量

                    continue;

                }

                else//can not get some one station,over

                {

                    printf("The maximum travel distance = %.2f\n", distance + Cmax*Davg);

                    break;

                }

            }

            if (Ncur > N)

            {

                printf("%.2lf\n", price);

            }

        }

    }

    return ;

}