#include <bits/stdc++.h>
using namespace std;
typedef struct {
  int c, m;
} Ele;
Ele m[3];
bool cmp(Ele a, Ele b) { return a.m < b.m; }
long long ans;
int main() {
  cin >> m[0].c >> m[1].c >> m[2].c;
  ans = m[0].c / 3 + m[1].c / 3 + m[2].c / 3;
  m[0].m = m[0].c % 3;
  m[1].m = m[1].c % 3;
  m[2].m = m[2].c % 3;
  if (m[0].m == m[1].m && m[1].m == m[2].m)
    ans += m[0].m;
  else {
    sort(m, m + 3, cmp);
    if ((m[0].c && m[0].m == 0 && m[1].m == 2 && m[2].m == 2) ||
        (m[0].m == 1 && m[1].m == 1 && m[2].m == 2) ||
        (m[0].m == 1 && m[1].m == 2 && m[2].m == 2))
      ans++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

Fox Ciel has some flowers: r red flowers, g green flowers and b blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:

  * To make a "red bouquet", it needs 3 red flowers. 
  * To make a "green bouquet", it needs 3 green flowers. 
  * To make a "blue bouquet", it needs 3 blue flowers. 
  * To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower. 



Help Fox Ciel to find the maximal number of bouquets she can make.

Input

The first line contains three integers r, g and b (0 ≤ r, g, b ≤ 109) — the number of red, green and blue flowers.

Output

Print the maximal number of bouquets Fox Ciel can make.

Examples

Input

3 6 9


Output

6


Input

4 4 4


Output

4


Input

0 0 0


Output

0

Note

In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.

In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef struct {
  int c, m;
} Ele;
Ele m[3];
bool cmp(Ele a, Ele b) { return a.m < b.m; }
long long ans;
int main() {
  cin >> m[0].c >> m[1].c >> m[2].c;
  ans = m[0].c / 3 + m[1].c / 3 + m[2].c / 3;
  m[0].m = m[0].c % 3;
  m[1].m = m[1].c % 3;
  m[2].m = m[2].c % 3;
  if (m[0].m == m[1].m && m[1].m == m[2].m)
    ans += m[0].m;
  else {
    sort(m, m + 3, cmp);
    if ((m[0].c && m[0].m == 0 && m[1].m == 2 && m[2].m == 2) ||
        (m[0].m == 1 && m[1].m == 1 && m[2].m == 2) ||
        (m[0].m == 1 && m[1].m == 2 && m[2].m == 2))
      ans++;
  }
  cout << ans << endl;
  return 0;
}

```