#include <bits/stdc++.h>
using namespace std;
const int N = 2000;
int ab[100000], ac[100000], bc[100000];
bool match(char x, char y){
    return x == '?' || y == '?' || x == y;
}
int main(){
    string a,b,c;
    cin >> a >> b >> c;
    int A = a.length();
    int B = b.length();
    int C = c.length();
    for (int i = 0; i < A; i++) for (int j = 0; j < B; j++) if (!match(a[i],b[j])) ab[i-j+50000] = true;
    for (int i = 0; i < A; i++) for (int j = 0; j < C; j++) if (!match(a[i],c[j])) ac[i-j+50000] = true;
    for (int i = 0; i < B; i++) for (int j = 0; j < C; j++) if (!match(b[i],c[j])) bc[i-j+50000] = true;
    int ans = 3*N;
    for (int i = -2*N; i <= 2*N; i++){
        for (int j = -2*N; j <= 2*N; j++){
            if (!ab[i+50000] && !ac[j+50000] && !bc[j-i+50000]){
                int L = min(0,min(i,j));
                int R = max(A,max(B+i,C+j));
                ans = min(ans,R-L);
            }
        }
    }
    printf("%d",ans);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

Examples

Input

a?c
der
cod


Output

7


Input

atcoder
atcoder
???????


Output

7

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2000;
int ab[100000], ac[100000], bc[100000];
bool match(char x, char y){
    return x == '?' || y == '?' || x == y;
}
int main(){
    string a,b,c;
    cin >> a >> b >> c;
    int A = a.length();
    int B = b.length();
    int C = c.length();
    for (int i = 0; i < A; i++) for (int j = 0; j < B; j++) if (!match(a[i],b[j])) ab[i-j+50000] = true;
    for (int i = 0; i < A; i++) for (int j = 0; j < C; j++) if (!match(a[i],c[j])) ac[i-j+50000] = true;
    for (int i = 0; i < B; i++) for (int j = 0; j < C; j++) if (!match(b[i],c[j])) bc[i-j+50000] = true;
    int ans = 3*N;
    for (int i = -2*N; i <= 2*N; i++){
        for (int j = -2*N; j <= 2*N; j++){
            if (!ab[i+50000] && !ac[j+50000] && !bc[j-i+50000]){
                int L = min(0,min(i,j));
                int R = max(A,max(B+i,C+j));
                ans = min(ans,R-L);
            }
        }
    }
    printf("%d",ans);
}

```