Problem:
Score : 500 points

Problem Statement

Snuke has a string s.
From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:

Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with ?s.

For example, if s is mississippi, we can choose the substring ssissip and replace its 1-st and 3-rd characters with ? to obtain ?s?ssip.

You are given the strings a, b, and c.
Find the minimum possible length of s.

Constraints

1 \leq |a|, |b|, |c| \leq 2000

a, b, and c consists of lowercase English letters and ?s.

Input

Input is given from Standard Input in the following format:

a
b
c

Output

Print the minimum possible length of s.

Sample Input 1

a?c
der
cod

Sample Output 1

7

For example, s could be atcoder.

Sample Input 2

atcoder
atcoder
???????

Sample Output 2

7

a, b, and c may not be distinct.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
 
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<ll, ll> pii;
 
#define REP(a,b,c) for(ll (a)=(b);(a)<(c);(a)++)
#define PER(a,b,c) for(ll (a)=(b);(a)>=(c);(a)--)
 
void ii(ll &x){ cin >> x; }
void is(string &x){ cin >> x; }
void oi(ll x){ cout << x << " "; }
void os(string x){ cout << x; }
void oie(ll x){ cout << x << endl; }
void ose(string x){ cout << x << endl; }
void oe(){ cout << endl; }

bool ab[10000],ac[10000],bc[10000];

bool match(char c1, char c2){
    return (c1=='?'||c2=='?'||c1==c2);
}

int main(){
    ll i,j;
    ll M=2000;
    string a,b,c;
    is(a);
    is(b);
    is(c);

    ll A = a.length();
    ll B = b.length();
    ll C = c.length();

    REP(i,0,A) REP(j,0,B) if(!match(a[i],b[j])) ab[i-j+5000] = true;
    REP(i,0,A) REP(j,0,C) if(!match(a[i],c[j])) ac[i-j+5000] = true;
    REP(i,0,B) REP(j,0,C) if(!match(b[i],c[j])) bc[i-j+5000] = true;

    ll ans=3*M;
    for(i=-2*M;i<=2*M;i++){
        for(j=-2*M;j<=2*M;j++){
            if(!ab[i+5000] && !ac[j+5000] && !bc[j-i+5000]){
                ll L = min((ll)0, min(i,j));
                ll R = max(A, max(B+i,C+j));
                ans=min(ans,R-L);
            }
        }
    }
    oie(ans);
    return 0;
}