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)

#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use text_io::*;

fn main() {
    let t: String = read!();

    let mut t = t.chars().collect::<Vec<char>>();
    t.reverse();

    let mut bef = 'P';
    let mut ans = vec![];
    for (i, &x) in t.iter().enumerate() {
        if x == '?' {
            match bef {
                'D' => {
                    if i == t.len() - 1 || t[i + 1] == 'D' {
                        ans.push('D');
                    } else {
                        bef = 'P';
                        ans.push('P');
                    }
                }
                'P' => {
                    bef = 'D';
                    ans.push('D');
                }
                _ => {}
            }
        } else {
            ans.push(x);
            bef = x;
        }
    }

    ans.reverse();

    for i in ans {
        print!("{}", i);
    }

    println!();
}