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: Accepted)

#include<iostream>
#include<algorithm>
#include<bits/stdc++.h>
using namespace std;
//#define mp make_pair
#define pb push_back
#define int long long
#define pi 3.141592653589793238462643383279502
#define M 998244353
#define F first
#define S second
#define pii pair<int,int>
#define rep(i,n) for(int i=0;i<n;i++)
bool isPowerOfTwo(int n)
{
   if(n==0)
   return false;

   return (ceil(log2(n)) == floor(log2(n)));
}
int power(int x, unsigned int y)
{
    int res = 1;

    x = x;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x)%M;

        y = y>>1;
        x = (x*x)%M;
    }
    return res%M;
}
bool isprime(int x)
{
      if(x<2)
        return false;
    for(int i=2;i*i<=x;i++)
     {
        if (x%i==0)
            return false;
    }
    return true;
}
vector<pair<int,int>>primeFactors(int n)
{
    vector<pair<int,int>>v;
    int x=0;
    while (n % 2 == 0)
    {
        x++;
        n = n/2;
    }
    if(x>0)
    v.pb({2,x});
    for (int i = 3; i <= sqrt(n); i = i + 2)
    {
        int y=0;
        while (n % i == 0)
        {
            y++;
            n = n/i;
        }
        if(y>0)
        v.pb({i,y});
    }
    if (n > 2)
        v.pb({n,1});
    return v;
}
int maxSubArraySum(int a[], int size)
{
    int max_so_far = INT_MIN, max_ending_here = 0;

    for (int i = 0; i < size; i++)
    {
        max_ending_here = max_ending_here + a[i];
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;

        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}
void solve()
{
string s;
cin>>s;
for(int i=0;i<s.size();i++)
{
    if(s[i]=='?')
        s[i]='D';
}
cout<<s<<endl;
}
signed main()
{ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int t;
//cin>>t;
t=1;
while(t--)
{

solve();

}
}