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)

// writer: w33z8kqrqk8zzzx33
#include <bits/stdc++.h>
using namespace std;

// begin fast read template by CYJian (source: https://www.luogu.com.cn/paste/i11c3ppx)

namespace io {
	const int __SIZE = (1 << 21) + 1;
	char ibuf[__SIZE], *iS, *iT, obuf[__SIZE], *oS = obuf, *oT = oS + __SIZE - 1, __c, qu[55]; int __f, qr, _eof;
	#define Gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, __SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	inline void flush () { fwrite (obuf, 1, oS - obuf, stdout), oS = obuf; }
	inline void gc (char &x) { x = Gc(); }
	inline void pc (char x) { *oS ++ = x; if (oS == oT) flush (); }
	inline void pstr (const char *s) { int __len = strlen(s); for (__f = 0; __f < __len; ++__f) pc (s[__f]); }
	inline void gstr (char *s) { for(__c = Gc(); __c < 32 || __c > 126 || __c == ' ';)  __c = Gc();
		for(; __c > 31 && __c < 127 && __c != ' '; ++s, __c = Gc()) *s = __c; *s = 0; }
	template <class I> inline bool gi (I &x) { _eof = 0;
		for (__f = 1, __c = Gc(); (__c < '0' || __c > '9') && !_eof; __c = Gc()) { if (__c == '-') __f = -1; _eof |= __c == EOF; }
		for (x = 0; __c <= '9' && __c >= '0' && !_eof; __c = Gc()) x = x * 10 + (__c & 15), _eof |= __c == EOF; x *= __f; return !_eof; }
	template <class I> inline void print (I x) { if (!x) pc ('0'); if (x < 0) pc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10; while (qr) pc (qu[qr --]); }
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
} using io::pc; using io::gc; using io::pstr; using io::gstr; using io::gi; using io::print;

// end fast read template by CYJian

#define iter(i, a, b) for(int i=(a); i<(b); i++)
#define rep(i, a) iter(i, 0, a)
#define rep1(i, a) iter(i, 1, (a)+1)
#define log(a) cerr<<"\033[32m[DEBUG] "<<#a<<'='<<(a)<<" @ line "<<__LINE__<<"\033[0m"<<endl
#define all(a) a.begin(), a.end()
#define fi first
#define se second
#define pb push_back
#define mp make_pair
 
using ll=long long;
using pii=pair<int, int>;
//#define int ll
const int MOD = 1000000007;

char buf[200005];

signed main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    gstr(buf);
	int n = strlen(buf);
	rep(i, n-1) if(buf[i] == '?' && buf[i+1] == 'D') buf[i] = 'P';
	rep(i, n-1) if(buf[i] == 'P' && buf[i+1] == '?') buf[i+1] = 'D';
	int cur = 0;
	for(int i=n-1; i>=0; i--) {
		if(buf[i] == '?') buf[i] = (cur % 2) ? 'P' : 'D', cur++;
		else cur = 0;
	}
	pstr(buf);
}