CodeForces

Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!

The computers bought for the room were different. Some of them had only USB ports, some — only PS/2 ports, and some had both options.

You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.

You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.

Input

The first line contains three integers a, b and c (0 ≤ a, b, c ≤ 10⁵)  — the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.

The next line contains one integer m (0 ≤ m ≤ 3·10⁵)  — the number of mouses in the price list.

The next m lines each describe another mouse. The i-th line contains first integer val_i (1 ≤ val_i ≤ 10⁹)  — the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.

Output

Output two integers separated by space — the number of equipped computers and the total cost of the mouses you will buy.

Example

Input

2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2

Output

3 14

题目大意： 一批电脑装鼠标，有的是usb接口有的是ps接口，有的ps和usb可以，然后给出一个单子上面是价格和鼠标类型， 每一种只能买一次， 问最少花费多少钱。 

思路：开始用的set函数，因为觉得每种只能用一次，不会有重复的，所以正好利用set自动排序，结果wa在十七发，应该是会有重复的吧。  后来改为用数组，两个数组分别存用usb的和ps的，然后排序选出必须usb和ps的，接着两个数组比较后存入都可以的鼠标。 挺暴力的一道题。

AC代码

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<string>
#include<cstdlib>
using namespace std;
#define LL long long
const int MaxN = 3e5 + 5;
int a, b, c;
LL m[MaxN], n[MaxN];
LL T,q = 0, p = 0;
char t[10];
LL sum = 0;
int main()
{
    int ans = 0;
    LL u;
    scanf("%d %d %d", &a, &b, &c);
    scanf("%I64d", &T);
    for(int i = 0; i < T; i++) {
        scanf("%I64d %s", &u, t);
        if(t[0] == 'U') 
            m[p++] = u;
        else n[q++] = u;
    }
    sort(m, m + p);
    sort(n, n + q);
    int v, w;
    for(v = 0; v < a; v++) {
        if(m[v] == 0) break;
        ans++;
        sum = sum + m[v];
    }
    for(w = 0; w < b; w++) {
        if(n[w] == 0) break;
        ans++;
        sum = sum + n[w];
    }
    for(int i = 0; i < c; i++) {
        if(m[v] <= n[w] && m[v] != 0) sum = sum + m[v++];
        else if(n[w] < m[v] && n[w] != 0) sum = sum + n[w++]; 
        else if(m[v] == 0 && n[w] != 0) sum = sum + n[w++];
        else if(n[w] == 0 && m[v] != 0) sum = sum + m[v++];
        else break;
        ans++;
    }
    printf("%d %I64d
",ans, sum);
}