#include <iostream>
#include <string>

using namespace std;

#define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)

#define M 2000
#define ADJUST 4000

bool ab[ADJUST*2],ac[ADJUST*2],bc[ADJUST*2];

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

int main(void){
int i,j;

string a,b,c;
cin >> a >> b >> c;

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

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

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

cout << ans << endl;

return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

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>

using namespace std;

#define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)

#define M 2000
#define ADJUST 4000

bool ab[ADJUST*2],ac[ADJUST*2],bc[ADJUST*2];

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

int main(void){
int i,j;

string a,b,c;
cin >> a >> b >> c;

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

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

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

cout << ans << endl;

return 0;
}
```