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: Accepted)

#![allow(unused_imports)]
#![allow(unused_variables)]
use proconio::input;
use proconio::marker::*;
use std::cmp::{min, max};
use std::collections::{HashSet, HashMap, BinaryHeap};

fn main() {
    input! {
        mut t: Chars,
    }
    let n = t.len();
    for i in 0..n - 1{
        if t[i] == 'P' && t[i + 1] == '?'{
            t[i + 1] = 'D';
        }else if t[i] == '?'{
            if t[i + 1] == '?'{
                t[i] = 'P';
                t[i + 1] = 'D';
            }else if t[i + 1] == 'P'{
                t[i] = 'D';
            }else if t[i + 1] == 'D'{
                t[i] = 'P';
            }
        }
    }
    if t[n - 1] == '?'{
        t[n - 1] = 'D';
    }
    for i in t.iter(){
        print!("{}", i);
    }
    println!();
}