Problem: For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.

We have a string T consisting of `P`, `D`, and `?`.

Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.

Constraints

* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.

Input

Input is given from Standard Input in the following format:


T


Output

Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.

Examples

Input

PD?D??P


Output

PDPDPDP


Input

P?P?


Output

PDPD

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class Main implements Runnable
{
//    static int dx[]={-1,0,1,0};
//    static int dy[]={0,-1,0,1};
    static long mod=1000000007;
    
    public static void main(String[] args) {        
        new Thread(null, new Main(), "", 200 * 1024 * 1024).start();        
    }

    public void run() {
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        
//        int n=sc.nextInt();
        char c[]=sc.getCharSet();
        for (int i = 0; i < c.length; i++) {
    		if(c[i]=='?') {
    	        out.print("D");				
			} else 
    	        out.print(c[i]);				
		}
        out.println();
     	out.flush();
    }
    
    static void ret(String ans) {
    	System.out.println(ans);
    	System.exit(0);
    }

    static void ret(long ans) {
    	System.out.println(ans);
    	System.exit(0);
    }
    
    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }
}

//class P {
//	int x,c;long s;
//	P(int  x, int c, long s) {
//	  this.x=x;
//	  this.c=c;
//	  this.s=s;
//	}
//}


class P implements Comparable<P>{
	int x,c;long s;
	P(int  x, int c, long s) {
	  this.x=x;
	  this.c=c;
	  this.s=s;
	}

    public int compareTo(P p){
        return Long.signum(s-p.s);
    }
}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString()
    {
        while( st==null || !st.hasMoreTokens() )
			try {
				st = new StringTokenizer(x.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
        return st.nextToken();
    }
    public int nextInt()
    {
        return Integer.parseInt(nextString());
    }
    public int[] nextIntArray(int size) {
        int r[] = new int[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextInt(); 
        }
        return r;
    }
    public long[] nextLongArray(int size) {
        long r[] = new long[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextLong(); 
        }
        return r;
    }
    public char[] getCharSet() {
        return this.nextString().toCharArray();
    }    
    public long nextLong()
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}