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;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;

#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define pb push_back

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (int i = (b); i >= (a); --i)
#define R0F(i, a) ROF(i, 0, a-1)
#define trav(x, a) for (auto& x: a)

const int MX  = 2e5+5;
const int MOD = 1e9+7;
const ll  INF = 1061109567; // 4557430888798830399LL
const ld  EPS = 1e-9;
const ld  PI  = acos(-1);

int main() {

    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    string T; cin >> T;

    string res = "";
    F0R(i,sz(T)) {
        if (T[i] == '?') {
            if (i-1 >= 0 && T[i-1] == 'P') res += "D";
            else {
                if (i+1 < sz(T) && (T[i+1] == 'D' || T[i+1] == '?')) {
                    res += "PD"; i++;
                } else res += "D";
            }
        } else res += T[i];
    }

    assert(sz(res) == sz(T));

    cout << res << "\n";

}