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>
using namespace std;
typedef long long ll;
#define REP(i, n) for(ll i=0; i<(ll)n; i++)
#define REPR(i, n) for(ll i=(ll)n-1; i>=0; i--)
#define FOR(i, m, n) for(ll i=(ll)m; i<(ll)n; i++)
#define FORR(i, m, n) for(ll i=(ll)m-1; i>=(ll)n; i--)
#define ALL(x) (x).begin(), (x).end()
#define ASC(v) sort(ALL(v));
#define DESC(v) sort(ALL(v), greater<int>());
#define UNIQ(a) sort(ALL(a));(a).resize(unique(ALL(a)) - a.begin());
#define PRINTD(x, d) cout << fixed << setprecision(d) << x << "\n";
#define DEBUG(x) cout<<#x<<": "<<x<<"\n"
const double PI = acos(-1);
ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }
bool isprime(int x){ int i; if(x<2)return 0; else if(x==2) return 1; if(x%2==0) return 0; for(i=3; i*i<=x; i+=2) if(x%i==0) return 0; return 1; }
int digsum(int n) { int r=0; while(n>0) { r+=n%10; n/=10; } return r; }
vector<string> split(string s, char p){vector<string>v;stringstream m(s);string b;while(getline(m,b,p))v.push_back(b);return v;}
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; }
struct nya{ nya(){ cin.tie(0); ios::sync_with_stdio(false); } }nyan;

int main(){
    string T; cin >> T;
    REP(i, T.size()) {
        char c;
        if (i==0 && T[i] == '?') {
            cout << T[i];
            continue;
        }
        if (T[i] == '?') {
            if (T[i-1] == 'D') c = 'P';
            else c = 'D';
        } else {
            c = T[i];
        }
        cout << c;
    }
    cout << "\n";
}