#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

#define BIAS 20000
bool ab[BIAS * 2];
bool bc[BIAS * 2];
bool ca[BIAS * 2];

void init(const string& S, const string& T, bool *M) {
  for (int i = 0; i < S.size(); i++) {
    for (int j = 0; j < T.size(); j++) {
      if (S[i] != '?' && T[j] != '?' && S[i] != T[j]) {
        M[i-j+BIAS] = true;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  string A, B, C;
  cin >> A >> B >> C;

  init(A, B, ab);
  init(B, C, bc);
  init(C, A, ca);
  int la = A.size();
  int lb = B.size();
  int lc = C.size();
  int a = 0;
  int ret = la + lb + lc;
  for (int b = -4000; b <= 4000; b++) {
    for (int c = -4000; c <= 4000; c++) {
      if (ab[b-a+BIAS] || bc[c-b+BIAS] || ca[a-c+BIAS]) continue;
      int L = min(a, min(b, c));
      int R = max(a + la, max(b + lb, c + lc));
      ret = min(ret, R - L);
    }
  }
  cout << ret << endl;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <iostream>
#include <string>
#include <algorithm>
using namespace std;

#define BIAS 20000
bool ab[BIAS * 2];
bool bc[BIAS * 2];
bool ca[BIAS * 2];

void init(const string& S, const string& T, bool *M) {
  for (int i = 0; i < S.size(); i++) {
    for (int j = 0; j < T.size(); j++) {
      if (S[i] != '?' && T[j] != '?' && S[i] != T[j]) {
        M[i-j+BIAS] = true;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  string A, B, C;
  cin >> A >> B >> C;

  init(A, B, ab);
  init(B, C, bc);
  init(C, A, ca);
  int la = A.size();
  int lb = B.size();
  int lc = C.size();
  int a = 0;
  int ret = la + lb + lc;
  for (int b = -4000; b <= 4000; b++) {
    for (int c = -4000; c <= 4000; c++) {
      if (ab[b-a+BIAS] || bc[c-b+BIAS] || ca[a-c+BIAS]) continue;
      int L = min(a, min(b, c));
      int R = max(a + la, max(b + lb, c + lc));
      ret = min(ret, R - L);
    }
  }
  cout << ret << endl;
}
```