#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=2e3+5;
const int M=2e4+5;
char a[N],b[N],c[N];
bool ab[M],ac[M],bc[M];
bool cal(char a,char b) {
	return a=='?'||b=='?'||a==b;
}
int main() {
    //freopen("in.txt","r",stdin);
	scanf("%s%s%s",a,b,c);
	int A=strlen(a),B=strlen(b),C=strlen(c);
	for(int i=0;i<A;i++) {
		for(int j=0;j<B;j++) {
			if(!cal(a[i],b[j])) ab[i-j+4000]=true;
		}
		for(int j=0;j<C;j++) {
			if(!cal(a[i],c[j])) ac[i-j+4000]=true;
		}
	}
	for(int i=0;i<B;i++) {
		for(int j=0;j<C;j++) {
			if(!cal(b[i],c[j])) bc[i-j+4000]=true;
		}
	}
	int ans=1e9;
	for(int i=-4000;i<=4000;i++) {
		for(int j=-4000;j<=4000;j++) {
			if(!ab[i+4000]&&!ac[j+4000]&&!bc[j-i+4000]) {
				int L=min(0,min(i,j)),R=max(A,max(B+i,C+j));
				ans=min(ans,R-L);
			}
		}
	}
	printf("%d\n",ans);
    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 <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=2e3+5;
const int M=2e4+5;
char a[N],b[N],c[N];
bool ab[M],ac[M],bc[M];
bool cal(char a,char b) {
	return a=='?'||b=='?'||a==b;
}
int main() {
    //freopen("in.txt","r",stdin);
	scanf("%s%s%s",a,b,c);
	int A=strlen(a),B=strlen(b),C=strlen(c);
	for(int i=0;i<A;i++) {
		for(int j=0;j<B;j++) {
			if(!cal(a[i],b[j])) ab[i-j+4000]=true;
		}
		for(int j=0;j<C;j++) {
			if(!cal(a[i],c[j])) ac[i-j+4000]=true;
		}
	}
	for(int i=0;i<B;i++) {
		for(int j=0;j<C;j++) {
			if(!cal(b[i],c[j])) bc[i-j+4000]=true;
		}
	}
	int ans=1e9;
	for(int i=-4000;i<=4000;i++) {
		for(int j=-4000;j<=4000;j++) {
			if(!ab[i+4000]&&!ac[j+4000]&&!bc[j-i+4000]) {
				int L=min(0,min(i,j)),R=max(A,max(B+i,C+j));
				ans=min(ans,R-L);
			}
		}
	}
	printf("%d\n",ans);
    return 0;
}
```