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 ll long long 
#define mod 998244353
#define MOD 998244353
ll fact[1000001];
void fast(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}
inline ll mul(ll a,ll b){
  a=(a+MOD)%MOD;
  b=(b+MOD)%MOD;
  ll c=a*b;
  c=(c+MOD)%MOD;
  return c;
}
inline ll add(ll a,ll b){
  a=(a+MOD)%MOD;
  b=(b+MOD)%MOD;
  ll ans=(a+b);
  ans=(ans+MOD)%MOD;
  return ans;
}

ll power(ll a,ll b,ll c){
  ll ans=1;
  while(b){
    if(b&1){
      ans=mul(ans,a);
    }
    a=mul(a,a);
    b=b/2;
  }
  return ans;
}
ll ncr(ll n,ll r){
    if(r>n)return 0;
    if(r==n)return 1;
    ll xx=fact[n];
    xx=xx*power(fact[r],mod-2,mod)%mod;
    xx=xx*power(fact[n-r],mod-2,mod)%mod;
    return xx;
}
int main() 
{ 
    fast();
    string s; cin>>s;
    string z="";
    int n=s.length();
    if(s[0]=='?')z.push_back('D');
    else z.push_back(s[0]);
    for(int i=1;i<n;++i){
        if(s[i]=='?'){
            z.push_back('D');
        }
        else z.push_back(s[i]);
    }
    cout<<z;
    return 0; 
}