#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<map>

#include<string>

using namespace std;

const int maxn = ;

struct Node

{

    int id,val;

};

int order[];

long long R[];

int S,T,N,M;

struct Myqueue

{

    Node Q[maxn];

    int head,tail;

    void init()

    {

        head = tail = ;

    }

    void push(int id,int val)

    {

        while(head < tail && Q[tail-].val+S*(id-Q[tail-].id) >= val)

            tail--;

        Q[tail].val = val;

        Q[tail++].id = id;

    }

    void pop(int id)

    {

        while(head < tail && Q[head].id < id - T)  //单调区间的最小值。

            head++;

    }

    Node getfront()

    {

        return Q[head];

    }

}solver;

map<string,int> mymap;

void init()

{

    mymap["Jan"] = ;  mymap["Feb"] = ;  mymap["Mar"] = ;

    mymap["Apr"] = ;  mymap["May"] = ;  mymap["Jun"] = ;

    mymap["Jul"] = ;  mymap["Aug"] = ;  mymap["Sep"] = ;

    mymap["Oct"] = ; mymap["Nov"] = ; mymap["Dec"] = ;

}

int mou[] = {,,,,,,,,,,,,};

void GetOrder()

{

    string s;

    int date,year,H;

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

    {

        cin>>s;

        scanf("%d %d %d %I64d",&date,&year,&H,&R[i]);

        int mouth = mymap[s];

        year -= ;

        order[i] = (*+)*(year/);

        year = year%;

        if(year == )

        {

            for(int j=; j<mouth; j++)

            {

                order[i] += mou[j];

                if(j == ) order[i] += ;

            }

        }

        else

        {

            order[i] +=  + (year-)*;

            for(int j=; j<mouth; j++)

                order[i] += mou[j];

        }

        order[i] += date - ;

        order[i] = order[i]* + H + ;

    }

}

int main()

{

    //freopen("E:\\acm\\input.txt","r",stdin);

    init();

    while(cin>>N>>M && N+M)

    {

        GetOrder();

        scanf("%d %d",&T,&S);

        solver.init();

        int cnt = ;

        long long ans = ;

        for(int i=; i<=M; i++)

        {

            int cost;

            scanf("%d",&cost);

            solver.push(i,cost);

            solver.pop(i);         //先去掉过期的

            Node cur = solver.getfront();

            while(order[cnt] == i && cnt <= N)  //cnt的上限，和while循环，考虑了，为啥不加。

            {

                ans += (cur.val + (i-cur.id)*S ) * R[cnt];  //乘法溢出问题

                cnt ++;

            }

        }

        printf("%I64d\n",ans);

    }

}