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;
#define rep(i,n) for(int i = 0; i < n; i++)
#define print(n) cout << n << endl
#define in(v) int v; cin >> v
#define vint vector<int>
#define pb push_back
typedef long long int ll;


int main(){
  string t;cin>>t;
  if(t.size()==1){
    if(t[0]=='?'){
      t[0]='D';
    }
    print(t);
    return 0;
  }
  for(int i=0;i<t.size()-2;i++){
    if(t[i+1]=='?'){
      if(t[i]=='P'){
        t[i+1]='D';
      }else if(t[i]=='D'){
        if(t[i+2]=='?'){
          t[i+2]='D';
          t[i+1]='P';
        }else if(t[i+2]=='D'){
          t[i+1]='P';
        }else{
          t[i+1]='D';
        }
      }else{
        t[i+1]='D';
      }
    }
  }
  if(t[0]=='?')t[0]='D';
  if(t[t.size()-1]=='?'){
    t[t.size()-1]='D';
  }
  print(t);
  return 0;
}