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: Accepted)

#include<stdio.h>

int main(void){

	char string[200001];
	char changedString[200001];

	int p = 'P', d = 'D', question = '?';
	int i, answerp =0, answerd =0;

	gets(string);

	//copystring
	for (i = 0;string[i] != '\0';i++) {
		changedString[i] = string[i];
	}
	changedString[i] = '\0';

	//changechar
	for (i = 0;string[i] != '\0';i++) {
		if (string[i] == question) {
			string[i] = p;
			changedString[i] = d;
		}
	}

	//count
	for (i = 1;string[i]!='\0';i++) {
		if (string[i] == d ) {
			answerp++;
			if (string[i - 1] == p) {
				answerp++;
			}
		}
	}

	for (i = 1;changedString[i] != '\0';i++) {
		if (changedString[i] == d) {
			answerd++;
			if (string[i - 1] == p) {
				answerd++;
			}
		}
	}

	if (answerp > answerd) {
		printf("%s", string);

	}
	else
	{
		printf("%s", changedString);

	}


	}