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 Rust (1.42.0) (judge verdict: Wrong Answer)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
prog_comp::init!();

unsafe fn solve(io: &mut Io) {
    let mut t = io.rc();

    let mut i = 0;
    while i < t.len() {
        match t[i] {
            '?' => {
                if i == 0 {
                    let mut j = i + 1;
                    while j < t.len() && t[j] == '?' {
                        j += 1;
                    }

                    t[i] = if j == t.len() {
                        'P'
                    } else if (j - i) % 2 == 0 {
                        t[j]
                    } else if t[j] == 'P' {
                        'D'
                    } else {
                        'P'
                    };
                } else if t[i - 1] == 'P' || i + 1 == t.len() || t[i + 1] == 'P' {
                    t[i] = 'D';
                } else {
                    t[i] = 'P';
                }
            }
            _ => {}
        }
        i += 1;
    }

    io.w(t.into_iter().collect::<String>());
}
*/

fn main() {
    let exe = "/tmp/binE31C8EC8";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";