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>
using namespace std;
// typedef ll long long;
#define REP(i,j,n) for(int i=j;i<n;i++)
#define rep(i,n) REP(i,0,n)



// Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main(){
    string s;
    cin >> s;
    int n = s.length();
    rep(i,n)
    {
        if (s[i]=='?')
        {
            // ここから数を数える
            int idx = i;
            while(idx < n)
            {
                if (s[idx] == '?') idx++;
                else break;
            }
            int hatena_sz = idx - i;
            
            // ここから書き込み
            // cout << hatena_sz << endl;
            bool flag;
            flag = false;
            if (i!=0 && s[i-1]=='P') flag = true;

            for(;i<idx;i++)
            {
                if (idx-i==1)
                {
                    s[i] = 'D';
                    break;
                }

                if (flag)
                {
                    s[i] = 'D';
                }
                else
                {
                    s[i] = 'P';
                }
                flag = !flag;
            }
            // cout << s << endl;
        }
    }
    cout << s << endl;
}