#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; (i) < (n); (i)++)
using namespace std;

long modpow(long a, long n, long mod) {
    long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

// a^{-1} mod を計算する
long modinv(long a, long mod) {
    return modpow(a, mod - 2, mod);
}

bool match(char a, char b){
    return (a == '?' || b == '?' || a == b);
}

int main()
{
    string a, b, c; cin >> a >> b >> c;
    int A = a.length();
    int B = b.length();
    int C = c.length();
    bool ab[100000], ac[100000], bc[100000];
    REP(i, 100000){
        ab[i] = true;
        ac[i] = true;
        bc[i] = true;
    }

    REP(i, A) REP(j, B) if(!match(a[i], b[j])) ab[i-j+50000] = false; 
    REP(i, A) REP(j, C) if(!match(a[i], c[j])) ac[i-j+50000] = false; 
    REP(i, B) REP(j, C) if(!match(b[i], c[j])) bc[i-j+50000] = false;

    int ans = 6000;
    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]){
                int L = min(0, min(i, j));
                int R = max(A, max(i+B, j+C));
                ans = min(ans , R-L);
            }
    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>
#define REP(i, n) for(int i = 0; (i) < (n); (i)++)
using namespace std;

long modpow(long a, long n, long mod) {
    long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

// a^{-1} mod を計算する
long modinv(long a, long mod) {
    return modpow(a, mod - 2, mod);
}

bool match(char a, char b){
    return (a == '?' || b == '?' || a == b);
}

int main()
{
    string a, b, c; cin >> a >> b >> c;
    int A = a.length();
    int B = b.length();
    int C = c.length();
    bool ab[100000], ac[100000], bc[100000];
    REP(i, 100000){
        ab[i] = true;
        ac[i] = true;
        bc[i] = true;
    }

    REP(i, A) REP(j, B) if(!match(a[i], b[j])) ab[i-j+50000] = false; 
    REP(i, A) REP(j, C) if(!match(a[i], c[j])) ac[i-j+50000] = false; 
    REP(i, B) REP(j, C) if(!match(b[i], c[j])) bc[i-j+50000] = false;

    int ans = 6000;
    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]){
                int L = min(0, min(i, j));
                int R = max(A, max(i+B, j+C));
                ans = min(ans , R-L);
            }
    cout << ans << endl;
    return 0;
    
}


```