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 PyPy3 (7.3.0) (judge verdict: Accepted)

# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import itertools
input = sys.stdin.readline
from collections import defaultdict
from heapq import heappop, heappush

from decimal import *

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

##### 累乗 #####
# pow(x, y, z) -> x**y % z

##### 割り算の切り上げ #####
# tmp = -(-4 // 3)

##### dict の for文 #####
# for k, v in d.items():
#     print(k, v)

##### 進数の変換 #####
# bin_str = bin(i)[2:]
# oct_str = oct(i)[2:]
# hex_str = hex(i)[2:]

def inputInt(): return int(input())
def inputMap(): return map(int, input().split())
def inputList(): return list(map(int, input().split()))
def inputStr(): return input()[:-1]

inf = float('inf')
mod = 1000000007

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

def main():
	T = inputStr()

	print(T.replace('?', 'D'))



#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# 2分検索のリバース
def bisect_left_reverse(a, x):
    if a[0] <= x:
        return 0
    if x < a[-1]:
        return len(a)
    # 二分探索
    ok = len(a) - 1
    ng = 0
    while (abs(ok - ng) > 1):
        mid = (ok + ng) // 2
        if a[mid] <= x:
            ok = mid
        else:
            ng = mid
    return ok

def bisect_right_reverse(a, x):
    if a[0] < x:
        return 0
    if x <= a[-1]:
        return len(a)
    # 二分探索
    ok = len(a) - 1
    ng = 0
    while (abs(ok - ng) > 1):
        mid = (ok + ng) // 2
        if a[mid] < x:
            ok = mid
        else:
            ng = mid
    return ok

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# 2分検索
# 存在しない場合: None 存在する場合: その値
def binary_search(list, item):
    low = 0
    high = len(list) - 1

    while low <= high:
        mid = (low + high) //2
        guess = list[mid]
        if guess == item:
            return mid
        if guess > item:
            high = mid -1
        else:
            low = mid + 1
    return None

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# N 個のボールを K グループに分ける場合のパターン数
def sunuke(N, K, mod=10**9+7):
	if N < K or K-1 < 0:
		return 0
	else:
		return combination(N-1, K-1, mod)

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nCr mod m
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10**9+7):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res

# mを法とするaの乗法的逆元
def modinv(a, mod=10**9+7):
    return pow(a, mod-2, mod)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nHr mod m
# 問題によって、combination()を切り替えること
def H(n, r, mod=10**9+7):
	# comb = Combination(n+r-1, mod)
	# return comb(n+r-1, r)
    return combination(n+r-1, r, mod)


class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * modinv[self.mod%i] % self.mod
        return modinv

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# dfs のサンプル
def dfs(graph,parent,counter,edge):
    stk = []
    stk.append(edge)
    while len(stk) > 0:
        p = stk.pop()
        for e in graph[p]:
            if parent[p] == e:
                continue
            else:
                parent[e] = p
                counter[e] += counter[p]
                stk.append(e)

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
if __name__ == "__main__":
	main()