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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
#define debug(x) cout << #x << " = " << x << '\n'
#define debug_arr(a , n) for(int i = 0 ; i < n ; i++)cout << a[i] << " "
#define speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define mp make_pair
#define pb push_back
#define ff first
#define ss second
#define vi vector<int>
#define vll vector<ll> 
#define inf 1000000000
#define mod 1000000007

const int max_n = 1e5 + 9;

typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
ll power(ll a , ll b)
{
    ll prod = 1;
    while(b)
    {
        if(b&1)
        prod = (prod*a)%mod;
        a = (a*a)%mod;
        b >>= 1;
    }
    return prod;
}
int main()
{
    string s;
    cin >> s;

    for(char c : s){
        if(c == '?')cout << "D";
        cout << c;
    }
    return 0;
}