Problem: There was an epidemic in Monstropolis and all monsters became sick. To recover, all monsters lined up in queue for an appointment to the only doctor in the city.

Soon, monsters became hungry and began to eat each other. 

One monster can eat other monster if its weight is strictly greater than the weight of the monster being eaten, and they stand in the queue next to each other. Monsters eat each other instantly. There are no monsters which are being eaten at the same moment. After the monster A eats the monster B, the weight of the monster A increases by the weight of the eaten monster B. In result of such eating the length of the queue decreases by one, all monsters after the eaten one step forward so that there is no empty places in the queue again. A monster can eat several monsters one after another. Initially there were n monsters in the queue, the i-th of which had weight ai.

For example, if weights are [1, 2, 2, 2, 1, 2] (in order of queue, monsters are numbered from 1 to 6 from left to right) then some of the options are:

  1. the first monster can't eat the second monster because a1 = 1 is not greater than a2 = 2; 
  2. the second monster can't eat the third monster because a2 = 2 is not greater than a3 = 2; 
  3. the second monster can't eat the fifth monster because they are not neighbors; 
  4. the second monster can eat the first monster, the queue will be transformed to [3, 2, 2, 1, 2]. 



After some time, someone said a good joke and all monsters recovered. At that moment there were k (k ≤ n) monsters in the queue, the j-th of which had weight bj. Both sequences (a and b) contain the weights of the monsters in the order from the first to the last.

You are required to provide one of the possible orders of eating monsters which led to the current queue, or to determine that this could not happen. Assume that the doctor didn't make any appointments while monsters were eating each other.

Input

The first line contains single integer n (1 ≤ n ≤ 500) — the number of monsters in the initial queue.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the initial weights of the monsters.

The third line contains single integer k (1 ≤ k ≤ n) — the number of monsters in the queue after the joke. 

The fourth line contains k integers b1, b2, ..., bk (1 ≤ bj ≤ 5·108) — the weights of the monsters after the joke. 

Monsters are listed in the order from the beginning of the queue to the end.

Output

In case if no actions could lead to the final queue, print "NO" (without quotes) in the only line. 

Otherwise print "YES" (without quotes) in the first line. In the next n - k lines print actions in the chronological order. In each line print x — the index number of the monster in the current queue which eats and, separated by space, the symbol 'L' if the monster which stays the x-th in the queue eats the monster in front of him, or 'R' if the monster which stays the x-th in the queue eats the monster behind him. After each eating the queue is enumerated again. 

When one monster eats another the queue decreases. If there are several answers, print any of them.

Examples

Input

6
1 2 2 2 1 2
2
5 5


Output

YES
2 L
1 R
4 L
3 L


Input

5
1 2 3 4 5
1
15


Output

YES
5 L
4 L
3 L
2 L


Input

5
1 1 1 3 3
3
2 1 6


Output

NO

Note

In the first example, initially there were n = 6 monsters, their weights are [1, 2, 2, 2, 1, 2] (in order of queue from the first monster to the last monster). The final queue should be [5, 5]. The following sequence of eatings leads to the final queue:

  * the second monster eats the monster to the left (i.e. the first monster), queue becomes [3, 2, 2, 1, 2]; 
  * the first monster (note, it was the second on the previous step) eats the monster to the right (i.e. the second monster), queue becomes [5, 2, 1, 2]; 
  * the fourth monster eats the mosnter to the left (i.e. the third monster), queue becomes [5, 2, 3]; 
  * the finally, the third monster eats the monster to the left (i.e. the second monster), queue becomes [5, 5]. 



Note that for each step the output contains numbers of the monsters in their current order in the queue.

Solution (in JAVA):

Solution:
import java.awt.event.InputEvent;
import java.util.*;

import org.omg.PortableInterceptor.INACTIVE;

public class CodeforcesRound378
{
	private static int monstersInitial[] = new int[501]; //The weights of the monsters
	private static int monstersFinal[] = new int[501]; // The final weights of the monsters 
	private static int initialNoMonsters, finalNoMonsters; // Initial and final number of monsters 
	private static Vector <Integer> resIndex = new Vector<Integer>();
	private static Vector resDirection = new Vector<>();
	
	private static void eatLeft(int start, int end)
	{
		
	}
	
	public static void main(String[] args)
	{
		Scanner input = new Scanner(System.in);
		//INPUT 
		initialNoMonsters = input.nextInt(); 
		for (int index = 0; index < initialNoMonsters; index++)
			monstersInitial[index] = input.nextInt(); 
		
		finalNoMonsters = input.nextInt(); 
		for (int index = 0; index < finalNoMonsters; index++)
			monstersFinal[index] = input.nextInt(); 
		
		//ALGORITHM STARTS 
		int currentPosition = 0; // Keeps our current position 
		
		//Go through each of the final weights of monsters 
		for (int i = 0; i < finalNoMonsters; i++)
		{
			//If we've used all initial monsters, it's an invalid test case 
			if (currentPosition == initialNoMonsters)
			{
				System.out.print("NO");
				return; 
			}
			
			//System.out.println("Current final monster " + i);
			
			int segmentWeight = 0; // Keeps the total weight of the segment 
			int previousPosition = currentPosition; // We need to know the start of the segment
			int segmentMaximum = -1; // Refreshes the maximum to -1
			boolean maximumExists = false; // Turns true if maximum is found 
			int indexMaximum = 0; // The index of the maximum element
			int kurov = -1;
			//Start from where we last stopped, calculate the sum of weights 
			//When we reach a sum > than the current final monster weight, stop
			boolean kur = false, penis = false; 
			for (; currentPosition < initialNoMonsters; currentPosition++)
			{
				segmentWeight += monstersInitial[currentPosition];
				
				/*System.out.println("index: " + currentPosition);
				System.out.println("kur: " + kur);
				System.out.println("penis: " + penis);*/
				
				//Check if we went over the desired weight 
				if (segmentWeight > monstersFinal[i])
				{
					System.out.print("NO");
					return;
				}
				
				//Check if maximum exists 
				if (monstersInitial[currentPosition] != segmentMaximum && segmentMaximum != -1)
					maximumExists = true;
				
				//Compare to maximum 
				if (monstersInitial[currentPosition] > segmentMaximum)
				{			
					
					if (segmentMaximum != -1)
						penis = true; 
					
					segmentMaximum = monstersInitial[currentPosition];
					indexMaximum = currentPosition; 
					
					kur = false; 
				}
				
				if (monstersInitial[currentPosition] < segmentMaximum)
					kur = true; 
				
				if (monstersInitial[currentPosition] == segmentMaximum && kur == false && penis == false)
				{
					//System.out.println("vlagam u na majka ti pickata");
					indexMaximum = currentPosition; 
				}
				
								
				/*System.out.println("Current weight: " + segmentWeight);
				System.out.println("Segment maximum: " + segmentMaximum);
				System.out.println("Maximum exists: " + maximumExists);
				System.out.println("Current position: " + currentPosition);
				System.out.println("Previous position: " + previousPosition);
				input.nextLine();*/
				
				// Check if we obtained the required value 
				if (segmentWeight == monstersFinal[i])
					break;
			}
			
			/*System.out.println("FOR LOOP TERMINATED!");
			
			System.out.println("Segment maximum: " + segmentMaximum);
			System.out.println("Maximum index: " + indexMaximum);
			System.out.println("Previous position: " + previousPosition);
			System.out.println("Current position: " + currentPosition);*/
			
			//System.out.println(indexMaximum);
			
			//If no maximum is found 
			if (maximumExists == false && currentPosition-previousPosition > 0)
			{
				System.out.print("NO");
				return; 
			}
			
			//If we reached the end of the monsters array 
			if (segmentWeight != monstersFinal[i])
			{
				System.out.print("NO");
				return;
			}
			
			//If there is only one element in the segment 
			if (currentPosition-previousPosition == 0) 
			{
				currentPosition++;
				continue; 
			}
			
			//Now, let's add all consecutive eatings to the results
			
			//CASE 1: The maximum is the leftmost position
			if (indexMaximum == previousPosition)
				for (int j = previousPosition; j < currentPosition; j++)
				{
					resIndex.addElement(i);
					resDirection.addElement('R');
					
				}
			
			//CASE 2: The maximum is the rightmost position 
			else if (indexMaximum == currentPosition)
				for (int j = currentPosition; j > previousPosition; j--)
				{
					resIndex.addElement(j - previousPosition + i);
					resDirection.addElement('L');
				}
			
			//CASE 3: The maximum is between the two 
			else 
			{
				if (monstersInitial[indexMaximum] == monstersInitial[indexMaximum - 1])
				{
					for (int j = indexMaximum; j < currentPosition; j++)
					{
						resIndex.addElement(indexMaximum - previousPosition + i);
						resDirection.addElement('R');
					}
				
					for (int j = indexMaximum; j > previousPosition; j--)
					{
						resIndex.addElement(j - previousPosition + i);
						resDirection.addElement('L');
					}
				}
				
					
				else 
				{
					for (int j = indexMaximum; j > previousPosition; j--)
					{
						resIndex.addElement(j - previousPosition + i);
						resDirection.addElement('L');
					}
					
					for (int j = 0; j < currentPosition - indexMaximum; j++)
					{
						resIndex.addElement(i);
						resDirection.addElement('R');
					}
				}
			}
			
			currentPosition++;
		}
		
		if (currentPosition != initialNoMonsters)
		{
			System.out.println("NO");
			return;
		}
		
		System.out.println("YES");
		for (int i = 0; i < resIndex.size(); i++)
			System.out.println((resIndex.elementAt(i)+1) + " " +  resDirection.elementAt(i));
	}
}