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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;

#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define sz(x) int(x.size())

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const double PI=3.14159265358979323846;

int mod = 1e9+7;
int INF = 1e9+7;
// const long long INF = 1LL<<60;


int main(){
  string str;
  cin >> str;
  reverse(str.begin(), str.end());

  rep(i, sz(str)){
    if(str[i] == '?'){
      if(i==0){
        str[i] = 'D';
      }else{
        if(str[i-1] == 'D'){
          str[i] = 'P';
        }else{
          str[i] = 'D';
        }
      }
    }
  }
  
  reverse(str.begin(), str.end());

  cout << str << endl;
  
  return 0;
}