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)

#include<stdio.h>

int main(){
    char s[200000];
    int i, sum = 0;
    scanf("%s", s);
    for(i=0; s[i+1] != '\0'; i++){
        switch(s[i+1]){
            case '?':
                if(s[i] == 'P'){ //P?
                    sum += 2;
                    s[i+1] = 'D';
                }
                else if(s[i] == 'D'){ //D?
                    sum += 2;
                    s[i+1] = 'D';
                }
                else{
                    sum += 2; //??
                    s[i+1] = 'D';
                }
                break;
            case 'P': //PP, DP, ?P
                break;
            case 'D':
                if(s[i] == 'P'){ //PD
                    sum += 1;
                }
                else if(s[i] == 'D'){ //DD
                    sum += 2; 
                }
                else sum += 2; //?D
                break;
        }
    }
    for (i = 0; s[i] != '\0'; i++){
        printf("%c", s[i]);
    }
    //printf("%d", sum);
    return 0;
}