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>
#define rep(i,s,n) for (int i=s;i<n;++i)
#define drep(i,s,n) for (int i=n;i>=s;--i)
#define all(x) (x).begin(),(x).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define print(x) cout<<(x)<<endl
using namespace std;
typedef long long ll;

int gcd(int a,int b)
{if (a%b == 0) return(b);
 else          return(gcd(b,a%b));}
int lcm(int a,int b)
{return a*b/gcd(a,b);}
void printv(vector<auto> v) {cout<<'[';rep(i,0,v.size()){cout<<v[i]<<',';}cout<<']'<<endl;}
void printvv(vector<vector<auto>> vv)
{
  cout<<'['<<endl;
  rep(q,0,vv.size()){cout<<'[';rep(p,0,vv.at(q).size())cout<<vv.at(q).at(p)<<',';cout <<']'<<endl;}
  cout<<']'<<endl;
}
void solve(); int main(){cin.tie(0);ios::sync_with_stdio(0);solve();}
int step_x[] = {-1,0,0,1};
int step_y[] = {0,-1,1,0};
int INF=10e5;
//------------------------------------------------------------
int N,M,ans=0;
string S;

void solve() {
  cin>>S;

  char pre;
  rep(i,0,S.size())
  {
    if(S[i]=='?')
    {
      S[i]='D';
    }

  }
  print(S);
  return;
}