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)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("filed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let t:String = read();
    let mut c:Vec<char> = t.chars().collect();
    let mut d = 0;
    let mut pd = 0;
    for i in 0..t.len() {
        if c[i] == '?' {
            if i == t.len()-1 {
                c[i] = 'D';
            } else {
                if pd >= d  && c[i+1] == 'D' || c[i+1] == '?' {
                    c[i] = 'P';
                } else {
                    c[i] = 'D';
                }
            }
        }

        if c[i] == 'D' {
            d += 1;
        } else {
            d = 0;
        }
        if i >0 && i < t.len() && t[i-1..i+1].contains("PD") {
            pd += 1;
        } else {
            pd = 0;
        }
    }
    println!("{}", c.iter().collect::<String>());
}