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;
using ll=long long;
const ll MOD=(ll)1e9+7;
//const ll MOD=(ll)998244353;
const ll INF=(ll)1e18+10;
const double PI = 2.0*acos(0);

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
int dy[]={-1,1,0,0};
int dx[]={0,0,-1,1};

void solve(int ca){
    string T; cin>>T;
    ll ans=0;
    if(T.length()==1){
        if(T[0]=='?') cout<<"D"<<endl;
        else cout<<T<<endl;
        return;
    }
    for (int i = 0; i < T.length(); ++i) {
        if(T[i]!='?') continue;
        if(i==0) T[i]='D';
        else if(i==T.length()-1) T[i]='D';
        else{
            if(T[i]=='P') T[i]='D';
            else T[i]='D';
        }
    }
    cout<<T<<endl;
}

//#define MULTI
int main() {
#ifdef MULTI
    int N; cin>>N;
    for (int i = 0; i < N; ++i) {
        solve(i+1);
    }
#else
    solve(0);
#endif
    return 0;
}