Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

PD?D??P

Sample Output 1

PDPDPDP

This string contains three occurrences of D and three occurrences of PD as contiguous substrings, so its doctoral and postdoctoral quotient is 6, which is the maximum doctoral and postdoctoral quotient of a string obtained by replacing each ? in T with P or D.

Sample Input 2

P?P?

Sample Output 2

PDPD

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

/*
 * main.c
 *
 *  Created on: 2020/03/28
 *      Author: family
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MAX(a,b) (a > b ? a : b)
#define MIN(a,b) (a > b ? b : a)
typedef long long int ll;
typedef unsigned long long int ull;

int sort_inc(const void *a, const void *b) { return (*(int*)a - *(int*)b);}
int sort_dec(const void* a, const void* b) { return (*(int*)b - *(int*)a);}
int sort_dec_ll(const void *a, const void *b) {
    ll da = *(ll*)a, db = *(ll*)b;    int val = 0;
    if(da > db) { val = -1; }
    else if (da == db) { val = 0; }
    else { val = 1; }
    return val;
}
int sort_inc_ll(const void *a, const void *b) {
    ll da = *(ll*)a, db = *(ll*)b;    int val = 0;
    if(da > db) { val = 1; }
    else if (da == db) { val = 0; }
    else { val = -1; }
    return val;
}
int sort_dic(const void *a, const void *b) {
    char *pa = (char *)a;    char *pb = (char *)b;    int i = 0, val = 0, N = 10;
    for (i = 0; i < N; i++) {
    	char da = pa[i], db = pb[i];
    	if (da == db) continue;
  		if (da > db) val = 1; else val = -1;
        break;
    }
    return val;
}

void yesno(int ans) {
	if (ans == 1) {	printf("Yes\n");
	} else {		printf("No\n");	}
	return;
}

void okng(int ans) {
	if (ans == 1) { printf("OK\n");
	} else { 		printf("NG\n");	}
	return;
}

char T[200001] = {0};
int main()
{
	scanf("%s", T);
	int i = 0;
	int len = strlen(T);
	for (i = len-1; i >= 0; i--) {
		if (T[i] == 'P') {
			//
		} else if (T[i] == 'D') {
			//
		} else {
			if (i == (len-1)) {
				T[i] = 'D';
			} else {
				if (T[i+1] == 'D') {
					T[i] = 'P';
				} else {
					T[i] = 'D';
				}
			}
		}
	}
	printf("%s\n", T);
	//yesno(ans);
	//okng(ans);
	return 0;
}