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 <bits/stdc++.h>
#define PI 3.1415926535897
using namespace std;
typedef long long ll;
const int INF = 1000000000;
const ll LINF = 1000000000000000000; //1e18
const double EPS = 1e-10;

int main(void)
{
    string T;
    cin >> T;

    if (T[0] == '?')
    {
        if (T[1] == 'D' || T[1] == '?')
        {
            T[0] = 'P';
        }
        else
        {
            T[0] = 'D';
        }
    }

    for (int i = 1; i < T.size() - 1; i++)
    {
        if (T[i] == '?')
        {
            if (T[i - 1] == 'P')
            {
                T[i] = 'D';
            }
            else if (T[i - 1] == 'D')
            {
                if (T[i + 1] == 'D')
                {
                    T[i] = 'P';
                }
                else if (T[i + 1] == 'P')
                {
                    T[i] = 'D';
                }
                else if (T[i + 1] == '?')
                {
                    T[i] = 'P';
                }
            }
        }
    }

    if (T[T.size() - 1] == '?')
    {
        T[T.size() - 1] = 'D';
    }

    for (int i = 0; i < T.size(); i++)
    {
        cout << T[i];
    }
    printf("\n");
    return (0);
}