#include<bits/stdc++.h>
using namespace std;
int main() {
	int ans=6000;
	string a,b,c;
	vector<bool>ab(16001,1),ac(16001,1),bc(16001,1);
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin>>a>>b>>c;
	for(int i=0;i<a.size();i++)
		for(int j=0;j<b.size();j++)
			if(a[i]!='?'&&b[j]!='?'&&a[i]!=b[j])
				ab[i-j+8000]=0;
	for(int i=0;i<a.size();i++)
		for(int j=0;j<c.size();j++)
			if(a[i]!='?'&&c[j]!='?'&&a[i]!=c[j])
				ac[i-j+8000]=0;
	for(int i=0;i<b.size();i++)
		for(int j=0;j<c.size();j++)
			if(b[i]!='?'&&c[j]!='?'&&b[i]!=c[j])
				bc[i-j+8000]=0;
	for(int i=-4000;i<=4000;i++)
		for(int j=-4000;j<=4000;j++)
			if(ab[i+8000]&&ac[j+8000]&&bc[j-i+8000])
				ans=min(ans,max((int)a.size(),max((int)b.size()+i,(int)c.size()+j))-min(0,min(i,j)));
	cout<<ans<<endl;
	return 0;
}
### Prompt

Create a solution in Cpp for 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;
int main() {
	int ans=6000;
	string a,b,c;
	vector<bool>ab(16001,1),ac(16001,1),bc(16001,1);
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin>>a>>b>>c;
	for(int i=0;i<a.size();i++)
		for(int j=0;j<b.size();j++)
			if(a[i]!='?'&&b[j]!='?'&&a[i]!=b[j])
				ab[i-j+8000]=0;
	for(int i=0;i<a.size();i++)
		for(int j=0;j<c.size();j++)
			if(a[i]!='?'&&c[j]!='?'&&a[i]!=c[j])
				ac[i-j+8000]=0;
	for(int i=0;i<b.size();i++)
		for(int j=0;j<c.size();j++)
			if(b[i]!='?'&&c[j]!='?'&&b[i]!=c[j])
				bc[i-j+8000]=0;
	for(int i=-4000;i<=4000;i++)
		for(int j=-4000;j<=4000;j++)
			if(ab[i+8000]&&ac[j+8000]&&bc[j-i+8000])
				ans=min(ans,max((int)a.size(),max((int)b.size()+i,(int)c.size()+j))-min(0,min(i,j)));
	cout<<ans<<endl;
	return 0;
}
```