#include<bits/stdc++.h>
using namespace std;
string a,b,c;
int ab[100000],ac[100000],bc[100000];
inline int ok(char x,char y){return (x=='?'||y=='?'||x==y);}
int main()
{
	cin>>a>>b>>c;
	int A=a.size(),B=b.size(),C=c.size();
	for(int i=0;i<A;i++)
		for(int j=0;j<B;j++)
			if(!ok(a[i],b[j]))ab[i-j+50000]=true;
	for(int i=0;i<A;i++)
		for(int j=0;j<C;j++)
			if(!ok(a[i],c[j]))ac[i-j+50000]=true;
	for(int i=0;i<B;i++)
		for(int j=0;j<C;j++)
			if(!ok(b[i],c[j]))bc[i-j+50000]=true;
	int ans=6000;
	for(int i=-4000;i<=4000;i++)
		for(int j=-4000;j<=4000;j++)
			if(!(ab[i+50000]||ac[j+50000]||bc[j-i+50000]))	
				ans=min(ans,max(A,max(B+i,C+j))-min(0,min(i,j)));
	cout<<ans<<endl;
	return 0;
}
### Prompt

Please create a solution in cpp 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<bits/stdc++.h>
using namespace std;
string a,b,c;
int ab[100000],ac[100000],bc[100000];
inline int ok(char x,char y){return (x=='?'||y=='?'||x==y);}
int main()
{
	cin>>a>>b>>c;
	int A=a.size(),B=b.size(),C=c.size();
	for(int i=0;i<A;i++)
		for(int j=0;j<B;j++)
			if(!ok(a[i],b[j]))ab[i-j+50000]=true;
	for(int i=0;i<A;i++)
		for(int j=0;j<C;j++)
			if(!ok(a[i],c[j]))ac[i-j+50000]=true;
	for(int i=0;i<B;i++)
		for(int j=0;j<C;j++)
			if(!ok(b[i],c[j]))bc[i-j+50000]=true;
	int ans=6000;
	for(int i=-4000;i<=4000;i++)
		for(int j=-4000;j<=4000;j++)
			if(!(ab[i+50000]||ac[j+50000]||bc[j-i+50000]))	
				ans=min(ans,max(A,max(B+i,C+j))-min(0,min(i,j)));
	cout<<ans<<endl;
	return 0;
}
```