#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
string a,b,c;
bool ab[100000],ac[100000],bc[100000];
bool judge(char a,char b){return (a=='?'||b=='?'||a==b);}
int main()
{
    cin>>a>>b>>c;
    int l1=a.size(),l2=b.size(),l3=c.size(),ans=6000;
    for(int i=0;i<l1;i++) for(int j=0;j<l2;j++) if(!judge(a[i],b[j])) ab[i-j+50000]=1;
    for(int i=0;i<l1;i++) for(int j=0;j<l3;j++) if(!judge(a[i],c[j])) ac[i-j+50000]=1;
    for(int i=0;i<l2;i++) for(int j=0;j<l3;j++) if(!judge(b[i],c[j])) bc[i-j+50000]=1;
    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(l1,max(l2+i,l3+j))-min(0,min(i,j)));
        }
    }
    cout<<ans;
    return 0;
}

### Prompt

In CPP, your task is to solve 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;
typedef long long ll;
string a,b,c;
bool ab[100000],ac[100000],bc[100000];
bool judge(char a,char b){return (a=='?'||b=='?'||a==b);}
int main()
{
    cin>>a>>b>>c;
    int l1=a.size(),l2=b.size(),l3=c.size(),ans=6000;
    for(int i=0;i<l1;i++) for(int j=0;j<l2;j++) if(!judge(a[i],b[j])) ab[i-j+50000]=1;
    for(int i=0;i<l1;i++) for(int j=0;j<l3;j++) if(!judge(a[i],c[j])) ac[i-j+50000]=1;
    for(int i=0;i<l2;i++) for(int j=0;j<l3;j++) if(!judge(b[i],c[j])) bc[i-j+50000]=1;
    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(l1,max(l2+i,l3+j))-min(0,min(i,j)));
        }
    }
    cout<<ans;
    return 0;
}

```