description: As their story unravels, a timeless tale is told once again...
  
  Shirahime, a friend of Mocha's, is keen on playing the music game Arcaea and sharing Mocha interesting puzzles to solve. This day, Shirahime comes up with a new simple puzzle and wants Mocha to solve them. However, these puzzles are too easy for Mocha to solve, so she wants you to solve them and tell her the answers. The puzzles are described as follow.
  
  There are n squares arranged in a row, and each of them can be painted either red or blue.
  
  Among these squares, some of them have been painted already, and the others are blank. You can decide which color to paint on each blank square.
  
  Some pairs of adjacent squares may have the same color, which is imperfect. We define the imperfectness as the number of pairs of adjacent squares that share the same color.
  
  For example, the imperfectness of "BRRRBBR" is 3, with "BB" occurred once and "RR" occurred twice.
  
  Your goal is to minimize the imperfectness and print out the colors of the squares after painting. 
  
  Input
  
  Each test contains multiple test cases. 
  
  The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Each test case consists of two lines.
  
  The first line of each test case contains an integer n (1≤ n≤ 100) — the length of the squares row.
  
  The second line of each test case contains a string s with length n, containing characters 'B', 'R' and '?'. Here 'B' stands for a blue square, 'R' for a red square, and '?' for a blank square.
  
  Output
  
  For each test case, print a line with a string only containing 'B' and 'R', the colors of the squares after painting, which imperfectness is minimized. If there are multiple solutions, print any of them.
  
  Example
  
  Input
  
  
  5
  7
  ?R???BR
  7
  ???R???
  1
  ?
  1
  B
  10
  ?R??RB??B?
  
  
  Output
  
  
  BRRBRBR
  BRBRBRB
  B
  B
  BRRBRBBRBR
  
  Note
  
  In the first test case, if the squares are painted "BRRBRBR", the imperfectness is 1 (since squares 2 and 3 have the same color), which is the minimum possible imperfectness.
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import static java.lang.Math.*;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.File;
  import java.io.PrintStream;
  import java.io.PrintWriter;
  import java.math.BigInteger;
  public class Main {
   
  	
  	/* 10^(7) = 1s.
  	 * ceilVal = (a+b-1) / b */
  	
  	static final int mod = 1000000007;
  	static final long temp = 998244353;
  	static final long MOD = 1000000007;
  	static final long M = (long)1e9+7;
   
  	static class Pair implements Comparable<Pair> 
  	{
  		int first, second;
  		
  		public Pair(int first, int second) 
  		{
  			this.first = first;
  			this.second = second;
  		}
  		
  		public int compareTo(Pair ob) 
  		{
  			return (int)(first - ob.first);
  		}
  	}
  	
  	
  	static class Tuple implements Comparable<Tuple>
  	{
  		int first, second,third;
  		public Tuple(int first, int second, int third)
  		{
  			this.first = first;
  			this.second = second;
  			this.third = third;
  		}
  		public int compareTo(Tuple o)
  		{
  			return (int)(o.third - this.third);
  		}
  	}
  	
  	public static class DSU
  	{
  		int[] parent;
  		int[] rank; //Size of the trees is used as the rank
  		public DSU(int n)
  		{
  			parent = new int[n];
  			rank = new int[n];
  			Arrays.fill(parent, -1);
  			Arrays.fill(rank, 1);
  		}
  		
  		public int find(int i) //finding through path compression
  		{
  			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
  		}
  		
  		public boolean union(int a, int b) //Union Find by Rank 
  		{
  			a = find(a);
  			b = find(b);
  			
  			if(a == b) return false; //if they are already connected we exit by returning false.
  			
  			// if a's parent is less than b's parent
  			if(rank[a] < rank[b])
  			{
  				//then move a under b
  				parent[a] = b;
  			}
  			//else if rank of j's parent is less than i's parent
  			else if(rank[a] > rank[b])
  			{
  				//then move b under a
  				parent[b] = a;
  			}
  			//if both have the same rank.
  			else
  			{
  				//move a under b (it doesnt matter if its the other way around.
  				parent[b] = a;
  				rank[a] = 1 + rank[a];
  			}
  			return true;
  		}	
  	}
  
  	static class Reader 
  	{
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) throws IOException {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		
  		long[] longReadArray(int n) throws IOException {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  	
  	public static int gcd(int a, int b)
  	{
  		if(b == 0)
  		 return a;
  		
  		else
  		return gcd(b,a%b);
  	} 
   
  		public static long lcm(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  	public static long LongGCD(long a, long b)
  	{
  		if(b == 0)
  			 return a;
  			
  			else
  			return LongGCD(b,a%b);
  	}
  	
  		public static long LongLCM(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  		//Count the number of coprime's upto N
  		public static long phi(long n)  //euler totient/phi function
  	    { 
  	       long ans = n;
  //	       for(long i = 2;i*i<=n;i++)
  //	       {
  //	    	   if(n%i == 0)
  //	    	   {
  //	    		   while(n%i == 0) n/=i;
  //	    		   ans -= (ans/i);
  //	    	   }
  //	       }
  //	       
  //	       if(n > 1)
  //	       {
  //	    	   ans -= (ans/n);
  //	       }
  	       
  	       for(long i = 2;i<=n;i++)
  	       {
  	    	   if(isPrime(i))
  	    	   {
  	    		   ans -= (ans/i);
  	    	   }
  	       }
  	       
  	       return ans;
  	       
  	    }	
  	
  	public static long fastPow(long x, long n) 
  	{
  		if(n == 0)
  			return 1;
  		else if(n%2 == 0)
  			return fastPow(x*x,n/2);
  		else
  			return x*fastPow(x*x,(n-1)/2);
  	}
  	
  	   public static long modPow(long x, long y, long p)
  	    {
  	        long res = 1;
  	        x = x % p;
  	 
  	        while (y > 0) {
  	            if (y % 2 == 1)
  	                res = (res * x) % p;
  	 
  	            y = y >> 1; 
  	            x = (x * x) % p;
  	        }
  	        return res;
  	    }
  	   
  	   static long modInverse(long n, long p)
  	    {
  	        return modPow(n, p - 2, p);
  	    }
  	 
  	    // Returns nCr % p using Fermat's little theorem.
  	    
  	    public static long nCrModP(long n, long r,long p)
  	    {
  	    	if (n<r) 
  	              return 0;
  	          
  	        if (r == 0)
  	            return 1;
  	 
  	        long[] fac = new long[(int)(n) + 1];
  	        fac[0] = 1;
  	 
  	        for (int i = 1; i <= n; i++)
  	            fac[i] = fac[i - 1] * i % p;
  	 
  	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
  	                % p * modInverse(fac[(int)(n - r)], p)
  	                % p)
  	            % p;
  	    }
  	    
  	    public static long fact(long n) 
  	    {
  	    	  long[] fac = new long[(int)(n) + 1];
  		        fac[0] = 1;
  		 
  		        for (long i = 1; i <= n; i++)
  		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
  		        
  		        return fac[(int)(n)];
  		}
  	    
  	    public static long nCr(long n, long k) 
  	    {
  			long ans = 1;
  			for(long i = 0;i<k;i++)
  			{
  				ans *= (n-i);
  				ans /= (i+1);
  			}
  			return ans;
  		}
   
  		//Modular Operations for Addition and Multiplication. 
  		   public static long perfomMod(long x)
  		   {
  		        return ((x%M + M)%M);
  		    }
  		   public static long addMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)+perfomMod(b));
  		    }
  		  
  		   public static long subMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)-perfomMod(b));
  		   }
  		   
  		   public static long mulMod(long  a, long b)
  		   {
  		        return perfomMod(perfomMod(a)*perfomMod(b));
  		   }
  		   
  		   public static boolean isPrime(long n) 
  			{
  				if(n == 1)
  				{
  					return false;
  				}
  				for(int i = 2;i*i<=n;i++)
  				{
  					if(n%i == 0)
  					{
  						return false;
  					}
  				}
  				return true;
  			}
  			
  			public static List<Long> SieveList(int n)
  			{
  				boolean prime[] = new boolean[(int)(n+1)];
  				Arrays.fill(prime, true);
  				
  				List<Long> l = new ArrayList<>();
  				for (long p = 2; p*p<=n; p++) 
  				{ 
  					if (prime[(int)(p)] == true) 
  					{ 	
  					    for(long i = p*p; i<=n; i += p) 
  						{
  						    prime[(int)(i)] = false; 
  						}	
  					} 
  				} 
  		 
  				for (long p = 2; p<=n; p++) 
  				{
  				    if (prime[(int)(p)] == true)
  				    {
  				       l.add(p); 
  				    }
  				}
  				
  				return l;
  			}
  
  			public static long log2(long n)
  		    {
  				 long ans = (long)(log(n)/log(2));
  				 return ans;
  			}
  			   
  			public static boolean isPow2(long n)
  			{
  				 return (n  != 0 && ((n & (n-1))) == 0);
  			}	
  		   
  		   public static boolean isSq(int x)  
  		   {
  			    long s = (long)Math.round(Math.sqrt(x));
  			    return s*s==x;
  		   }
  		
  			   /*
  			    * 
  			    * >=         <=
  			   0 1 2 3 4 5 6 7  
  			   5 5 5 6 6 6 7 7
  			   lower_bound for 6 at index 3 (>=)
  			   upper_bound for 6 at index 6(To get six reduce by one) (<=)
  			   */
  
  		   public static int LowerBound(long a[], long x) 
  		   {
  			      int l=-1,r=a.length;
  				  while(l+1<r) 
  				  {
  				    int m=(l+r)>>>1;
  				    if(a[m]>=x) r=m;
  				    else l=m;
  				  }
  				  return r;
  			}
  		   
  			public static int UpperBound(long a[], long x) 
  			{
  				int l=-1, r=a.length;
  			    while(l+1<r) 
  			    {
  			       int m=(l+r)>>>1;
  			       if(a[m]<=x) l=m;
  			       else r=m;
  			    }
  			    return l+1;
  			}
  			
  			public static void Sort(long[] a) 
  			{
  				List<Long> l = new ArrayList<>();
  				for (long i : a) l.add(i);
  				Collections.sort(l);
  //				Collections.reverse(l);  //Use to Sort decreasingly 
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  			
  			public static void ssort(char[] a)
  			{
  				List<Character> l = new ArrayList<>();
  				for (char i : a) l.add(i);
  				Collections.sort(l);
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  	
  
  	
  	public static void main(String[] args) throws Exception 
  	{
  		Reader sc = new Reader();
  		PrintWriter fout = new PrintWriter(System.out);
  		
  		int t = sc.nextInt();
  		while(t-- > 0)
  		{
  			int n = sc.nextInt();
  			char[] s = sc.next().toCharArray();
  			
  			int idx = -1;
  			
  			for(int i = 0;i<n;i++)
  			{
  				if(s[i] == 'B') idx = 0;
  				else if(s[i] == 'R') idx = 1;
  				else if(idx == -1) continue;
  				else if(idx == 0)
  				{
  					s[i] = 'R';
  					idx = 1;
  				}
  				else if(idx == 1)
  				{
  					s[i] = 'B';
  					idx = 0;
  				}
  			}
  			
  			idx = 0;
  			for(int i = n-1;i>=0;i--)
  			{
  				if(s[i] == 'B') idx = 0;
  				else if(s[i] == 'R') idx = 1;
  				else if(idx == 0)
  				{
  					s[i] = 'R';
  					idx = 1;
  				}
  				else if(idx == 1)
  				{
  					s[i] = 'B';
  					idx = 0;
  				}
  			}
  			
  			fout.println(new String(s));
  		}
  
  		fout.close();
  	}
  }