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++ (Clang 10.0.0) (judge verdict: Runtime Error)

#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <utility>
#include <vector>

using namespace std;

void printString(string &t, int size) {
    for(int i = 0; i < size; i++) {
        cout << t[i] << "";
    }
    cout << endl;
}

int cntVector(string &t, long long t_size) {
    long long cnt = 0;
    for(int i = 0; i < t_size; i++) {
        if(i > 0 && t[i - 1] == 'P' && t[i] == '?') {
            t[i] = 'D';
            cnt++;
        }
        if(i > 0 && t[i - 1] == 'D' && t[i] == '?') {
            if(i + 1 == 'D' || i + 1 == '?') {
                t[i] = 'P';
            } else {
                t[i] = 'D';
                cnt++;
            }
            // (t[i + 1] == 'D') t[i] = 'P';
        }
        if(t[i] == 'D')
            cnt++;
    }
    return cnt;
}

int main() {
    string t, t1, t2;
    cin >> t;
    long long ma = 0;
    for(int i = 0; i < t.size(); i++) {
        t1[i] = t[i];
        t2[i] = t[i];
    }

    if(t[0] == '?') {
        t1[0] = 'P';
        t2[0] = 'D';
    }

    long long t_cnt = cntVector(t, t.size());
    long long t1_cnt = cntVector(t1, t.size());
    long long t2_cnt = cntVector(t2, t.size());

    ma = max(ma, t_cnt);
    ma = max(ma, t1_cnt);
    ma = max(ma, t2_cnt);

    if(t_cnt == ma)
        printString(t, t.size());
    else if(t1_cnt == ma)
        printString(t1, t.size());
    else
        printString(t2, t.size());

    return 0;
}