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>
#include <math.h>


    using namespace std;
    typedef long long ll;
    # define M_PI  3.14159265358979323846
    const int MOD=1e9+7;

    void solve(){
        string s;
        cin>>s;
        int n=s.size();
        for(int i=0;i<n;i++){
            if(s[i]=='?' && (i+1<n && s[i+1]=='P') || (i-1>=0 && s[i-1]=='P'))
                s[i]='D';
            else if(s[i]=='?' && i+1<n && s[i+1]=='?'){
                s[i]='P';
                s[i+1]='D';
            }
            else if(s[i]=='?' && i+1<n && s[i+1]=='D')
                s[i]='P';
            else if(s[i]=='?')
                s[i]='D';
        }
        cout<<s;
    }

    int main(){
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
        cout<<fixed;
        cout<<setprecision(10);
        int t=1;
//        cin>>t;
        for(int i=1;i<=t;i++){
            solve();
        }
        return 0;
    }