Problem: 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 vali (1 ≤ vali ≤ 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

Note

In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Scanner;
import java.util.Collections;

public class CF762B{
	public static void  main(String[] args){
		Scanner in = new Scanner(System.in);
		int a = in.nextInt(); // num of coms with usb
		int b = in.nextInt(); // num of coms with ps/2
		int c = in.nextInt(); // num of coms with both

		int m = in.nextInt(); // num of mouses

		// store the prices and type for m number of mouse using ArrayList(java data structure)
		ArrayList<Integer> usbPrices = new ArrayList<Integer>();
		ArrayList<Integer> ps2Prices = new ArrayList<Integer>(); 

		for(int i = 0; i < m; ++i){
			int price = in.nextInt();
			String type = in.next();

			// if(type.equals("USB")){
			// 	usbPrices.add(price);
			// } else if(type.equals("PS/2")){
			// 	ps2Prices.add(price);
			// }
			if(type.charAt(0) == 'U'){
				usbPrices.add(price);
			} else{
				ps2Prices.add(price);
			}
		}

		// sort the ArrayLists
		Collections.sort(usbPrices);
		Collections.sort(ps2Prices);

		// counter for number of computers got connected mouse
		int count = 0;

		// total price needs
		long total = 0;

		// indexing for the ArrayLists
		int usbIndex = 0;
		int ps2Index = 0;


		// add value to total and count for usb computers
		while(usbIndex < usbPrices.size() && usbIndex < a){
			total += usbPrices.get(usbIndex);
			++count;
			++usbIndex;
		}

		// add value to total and count for ps2 computers
		while(ps2Index < ps2Prices.size() && ps2Index < b){
			total += ps2Prices.get(ps2Index);
			++count;
			++ps2Index;
		}

		// add value to total and count for both 
		while(c > 0 && (usbIndex < usbPrices.size() || ps2Index < ps2Prices.size())){
			if(usbIndex == usbPrices.size()){
				total += ps2Prices.get(ps2Index);
				++ps2Index;
			} else if(ps2Index == ps2Prices.size()){
				total += usbPrices.get(usbIndex);
				++usbIndex;
			} else{
				if(usbPrices.get(usbIndex) < ps2Prices.get(ps2Index)){
					total += usbPrices.get(usbIndex);
					++usbIndex;
				} else{
					total += ps2Prices.get(ps2Index);
					++ps2Index;
				}
			}
			++count;
			--c;
		}

		// print output
		System.out.println(count + " " + total);
	}
}