One day, Hongcow goes to the store and sees a brand new deck of n special cards. Each individual card is either red or blue. He decides he wants to buy them immediately. To do this, he needs to play a game with the owner of the store.

This game takes some number of turns to complete. On a turn, Hongcow may do one of two things: 

  * Collect tokens. Hongcow collects 1 red token and 1 blue token by choosing this option (thus, 2 tokens in total per one operation). 
  * Buy a card. Hongcow chooses some card and spends tokens to purchase it as specified below. 



The i-th card requires ri red resources and bi blue resources. Suppose Hongcow currently has A red cards and B blue cards. Then, the i-th card will require Hongcow to spend max(ri - A, 0) red tokens, and max(bi - B, 0) blue tokens. Note, only tokens disappear, but the cards stay with Hongcow forever. Each card can be bought only once.

Given a description of the cards and their costs determine the minimum number of turns Hongcow needs to purchase all cards.

Input

The first line of input will contain a single integer n (1 ≤ n ≤ 16).

The next n lines of input will contain three tokens ci, ri and bi. ci will be 'R' or 'B', denoting the color of the card as red or blue. ri will be an integer denoting the amount of red resources required to obtain the card, and bi will be an integer denoting the amount of blue resources required to obtain the card (0 ≤ ri, bi ≤ 107).

Output

Output a single integer, denoting the minimum number of turns needed to acquire all the cards.

Examples

Input

3
R 0 1
B 1 0
R 1 1


Output

4


Input

3
R 3 0
R 2 0
R 1 0


Output

6

Note

For the first sample, Hongcow's four moves are as follows: 

  1. Collect tokens 
  2. Buy card 1
  3. Buy card 2
  4. Buy card 3

Note, at the fourth step, Hongcow is able to buy card 3 because Hongcow already has one red and one blue card, so we don't need to collect tokens.

For the second sample, one optimal strategy is as follows: 

  1. Collect tokens 
  2. Collect tokens 
  3. Buy card 2
  4. Collect tokens 
  5. Buy card 3
  6. Buy card 1

At the fifth step, even though Hongcow has a red token, Hongcow doesn't actually need to spend it, since Hongcow has a red card already.
import java.io.*;
import java.util.*;

public class ProblemE {
	
	public static int[][][] dp = new int[65536][2][257];
	public static int n = -1;
	public static ArrayList<Card> modCards = new ArrayList<Card>();
	public static int goal = -1;

	public static void main(String[] args) throws Exception {
		for(int i=0; i<65536; i++) {
			for(int j=0; j<257; j++) {
				dp[i][0][j] = -1;
				dp[i][1][j] = -1;
			}
		}
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		n = Integer.parseInt(br.readLine());
		goal = (1 << n) - 1;
		
		String[][] cards = new String[n][3];
		for(int i=0; i<n; i++) {
			cards[i] = br.readLine().split(" ");
		}
		
		int minRedCost = 0;
		int minBlueCost = 0;
		
		for(int i=0; i<n; i++) {
			boolean isRed;
			int redCost;
			int blueCost;
			if(cards[i][0].equals("R"))
				isRed = true;
			else
				isRed = false;
			redCost = Integer.parseInt(cards[i][1]);
			blueCost = Integer.parseInt(cards[i][2]);
			
			if(redCost > n) {
				minRedCost += (redCost - n);
				redCost = n;
			}
			if(blueCost > n) {
				minBlueCost += (blueCost - n);
				blueCost = n;
			}
			
			modCards.add(new Card(isRed, redCost, blueCost));
		}
		
		long minTurns;
		long sol;
		
		if(minRedCost > minBlueCost) {
			minTurns = minRedCost;
			int otherAmount = Math.min(minRedCost - minBlueCost, 256);
			sol = solve(0, 1, otherAmount);
		} else {
			minTurns = minBlueCost;
			int otherAmount = Math.min(minBlueCost - minRedCost, 256);
			sol = solve(0, 0, otherAmount);
		}
		
		System.out.println(minTurns + sol);
	}
	
	public static int solve(int mask, int redIsZero, int otherCount) {
		if(dp[mask][redIsZero][otherCount] != -1)
			return dp[mask][redIsZero][otherCount];
		
		if(mask == goal)
			return 0;
		
		int selector = 1;
		int redDiscount = 0;
		int blueDiscount = 0;
		for(int i=0; i<n; i++) {
			if((mask & selector) > 0) {
				if(modCards.get(i).isRed)
					redDiscount++;
				else
					blueDiscount++;
			}
			selector = selector << 1;
		}
		
		int min = 1000000007;
		selector = 1;
		for(int i=0; i<n; i++) {
			if((mask & selector) == 0) {
				int cand;
				int redPrice = Math.max(modCards.get(i).redCost - redDiscount, 0);
				int bluePrice = Math.max(modCards.get(i).blueCost - blueDiscount, 0);
				int turns = 1;
				int curRed;
				int curBlue;
				if(redIsZero == 1) {
					curRed = 0;
					curBlue = otherCount;
				} else {
					curRed = otherCount;
					curBlue = 0;
				}
				
				if(redPrice - curRed > bluePrice - curBlue) {
					turns += redPrice - curRed;
					int remBlue = Math.min(curBlue + redPrice - curRed - bluePrice, 256);
					cand = turns + solve(mask | selector, 1, remBlue);
				} else {
					turns += bluePrice - curBlue;
					int remRed = Math.min(curRed + bluePrice - curBlue - redPrice, 256);
					cand = turns + solve(mask | selector, 0, remRed);
				}
				if(cand < min)
					min = cand;	
			}		
			selector = selector << 1;
		}
		dp[mask][redIsZero][otherCount] = min;
		return dp[mask][redIsZero][otherCount];
	}

}

class Card {
	public boolean isRed;
	public int redCost;
	public int blueCost;
	public Card(boolean col, int red, int blue) {
		isRed = col;
		redCost = red;
		blueCost = blue;
	}
}