#include <iostream>
#include <cmath>
#include <complex>
#include <string>
#include <sstream>
#include <limits>
#include <algorithm>
#include <functional>

#include <vector>

using namespace std;

const int M = 2000;

vector<bool> bad_ab(16001, false);
vector<bool> bad_ac(16001, false);
vector<bool> bad_bc(16001, false);

bool match(char c1, char c2)
{
	return (c1 == '?' || c2 == '?' || c1 == c2);
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

	string a;
	string b;
	string c;

	cin >> a;
	cin >> b;
	cin >> c;

	int A = a.size();
	int B = b.size();
	int C = c.size();

	for (int i = 0; i < A; i++)
	{
		for (int j = 0; j < B; j++)
		{
			if (!match(a[i], b[j])) bad_ab[i - j + 8000] = true;
		}
	}

	for (int i = 0; i < A; i++)
	{
		for (int j = 0; j < C; j++)
		{
			if (!match(a[i], c[j])) bad_ac[i - j + 8000] = true;
		}
	}

	for (int i = 0; i < B; i++)
	{
		for (int j = 0; j < C; j++)
		{
			if (!match(b[i], c[j])) bad_bc[i - j + 8000] = true;
		}
	}

	int result = 3 * M;
	for (int i = -2 * M; i <= 2 * M; i++)
	{
		for (int j = -2 * M; j <= 2 * M; j++)
		{
			if (!bad_ab[i + 8000] && !bad_ac[j + 8000] && !bad_bc[j - i + 8000])
			{
				int L = min(0, min(i, j));
				int R = max(A, max(B + i, C + j));
				result = min(result, R - L);
			}
		}
	}

	cout << result;

	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 <iostream>
#include <cmath>
#include <complex>
#include <string>
#include <sstream>
#include <limits>
#include <algorithm>
#include <functional>

#include <vector>

using namespace std;

const int M = 2000;

vector<bool> bad_ab(16001, false);
vector<bool> bad_ac(16001, false);
vector<bool> bad_bc(16001, false);

bool match(char c1, char c2)
{
	return (c1 == '?' || c2 == '?' || c1 == c2);
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

	string a;
	string b;
	string c;

	cin >> a;
	cin >> b;
	cin >> c;

	int A = a.size();
	int B = b.size();
	int C = c.size();

	for (int i = 0; i < A; i++)
	{
		for (int j = 0; j < B; j++)
		{
			if (!match(a[i], b[j])) bad_ab[i - j + 8000] = true;
		}
	}

	for (int i = 0; i < A; i++)
	{
		for (int j = 0; j < C; j++)
		{
			if (!match(a[i], c[j])) bad_ac[i - j + 8000] = true;
		}
	}

	for (int i = 0; i < B; i++)
	{
		for (int j = 0; j < C; j++)
		{
			if (!match(b[i], c[j])) bad_bc[i - j + 8000] = true;
		}
	}

	int result = 3 * M;
	for (int i = -2 * M; i <= 2 * M; i++)
	{
		for (int j = -2 * M; j <= 2 * M; j++)
		{
			if (!bad_ab[i + 8000] && !bad_ac[j + 8000] && !bad_bc[j - i + 8000])
			{
				int L = min(0, min(i, j));
				int R = max(A, max(B + i, C + j));
				result = min(result, R - L);
			}
		}
	}

	cout << result;

	return 0;
}
```