#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;
bool ab[100000],bc[100000],ac[100000];
bool ok(char a,char b){
	return (!(a=='?'||b=='?'||a==b));
}
int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    for(int i=0;i<100000;i++)ab[i]=true;
    for(int i=0;i<100000;i++)bc[i]=true;
    for(int i=0;i<100000;i++)ac[i]=true;
    string a,b,c;
    cin>>a>>b>>c;
    int A=a.length(),B=b.length(),C=c.length();
    for(int i=0;i<A;i++){
    	for(int j=0;j<B;j++){
    		if(ok(a[i],b[j]))ab[i-j+50000]=false;
    	}
    }
    for(int i=0;i<A;i++){
    	for(int j=0;j<C;j++){
    		if(ok(a[i],c[j]))ac[i-j+50000]=false;
    	}
    }
    for(int i=0;i<B;i++){
    	for(int j=0;j<C;j++){
    		if(ok(b[i],c[j]))bc[i-j+50000]=false;
    	}
    }
    int m=2000;
    int ans=m*3;
    for(int i=-2*m;i<=2*m;i++){
    	for(int j=-2*m;j<=2*m;j++){
    		if(ab[i+50000]&&bc[j-i+50000]&&ac[j+50000]){
    			ans=min(ans,max(A,max(B+i,C+j))-min(0,min(i,j)));
    		}
    	}
    }
    cout << ans<<endl;
    return 0;
}

### Prompt

Generate 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 <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;
bool ab[100000],bc[100000],ac[100000];
bool ok(char a,char b){
	return (!(a=='?'||b=='?'||a==b));
}
int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    for(int i=0;i<100000;i++)ab[i]=true;
    for(int i=0;i<100000;i++)bc[i]=true;
    for(int i=0;i<100000;i++)ac[i]=true;
    string a,b,c;
    cin>>a>>b>>c;
    int A=a.length(),B=b.length(),C=c.length();
    for(int i=0;i<A;i++){
    	for(int j=0;j<B;j++){
    		if(ok(a[i],b[j]))ab[i-j+50000]=false;
    	}
    }
    for(int i=0;i<A;i++){
    	for(int j=0;j<C;j++){
    		if(ok(a[i],c[j]))ac[i-j+50000]=false;
    	}
    }
    for(int i=0;i<B;i++){
    	for(int j=0;j<C;j++){
    		if(ok(b[i],c[j]))bc[i-j+50000]=false;
    	}
    }
    int m=2000;
    int ans=m*3;
    for(int i=-2*m;i<=2*m;i++){
    	for(int j=-2*m;j<=2*m;j++){
    		if(ab[i+50000]&&bc[j-i+50000]&&ac[j+50000]){
    			ans=min(ans,max(A,max(B+i,C+j))-min(0,min(i,j)));
    		}
    	}
    }
    cout << ans<<endl;
    return 0;
}

```