#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, c;
  cin >> a >> b >> c;
  long long mod1 = a % 3;
  long long mod2 = b % 3;
  long long mod3 = c % 3;
  long long total = 0;
  if (a < 3 && b < 3 && c < 3) {
    total = 0;
  } else if (mod1 == mod2 && mod1 == mod3 && mod2 == mod3) {
    total += mod1;
    a -= mod1;
    b -= mod1;
    c -= mod1;
    total += a / 3 + b / 3 + c / 3;
  } else if (mod1 == mod2) {
    if (a >= 3 && b >= 3 && c >= 3) {
      total += mod1;
      a -= mod1;
      b -= mod1;
      c -= mod1;
    }
    total += a / 3 + b / 3 + c / 3;
  } else if (mod2 == mod3) {
    if (a >= 3 && b >= 3 && c >= 3) {
      total += mod2;
      a -= mod2;
      b -= mod2;
      c -= mod2;
    }
    total += (a / 3 + b / 3 + c / 3);
  } else if (mod1 == mod3) {
    if (a >= 3 && b >= 3 && c >= 3) {
      total += mod1;
      a -= mod1;
      b -= mod1;
      c -= mod1;
    }
    total += a / 3 + b / 3 + c / 3;
  } else {
    total += a / 3 + b / 3 + c / 3;
  }
  cout << total << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
const int mod = 1000000007;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, c;
  cin >> a >> b >> c;
  long long mod1 = a % 3;
  long long mod2 = b % 3;
  long long mod3 = c % 3;
  long long total = 0;
  if (a < 3 && b < 3 && c < 3) {
    total = 0;
  } else if (mod1 == mod2 && mod1 == mod3 && mod2 == mod3) {
    total += mod1;
    a -= mod1;
    b -= mod1;
    c -= mod1;
    total += a / 3 + b / 3 + c / 3;
  } else if (mod1 == mod2) {
    if (a >= 3 && b >= 3 && c >= 3) {
      total += mod1;
      a -= mod1;
      b -= mod1;
      c -= mod1;
    }
    total += a / 3 + b / 3 + c / 3;
  } else if (mod2 == mod3) {
    if (a >= 3 && b >= 3 && c >= 3) {
      total += mod2;
      a -= mod2;
      b -= mod2;
      c -= mod2;
    }
    total += (a / 3 + b / 3 + c / 3);
  } else if (mod1 == mod3) {
    if (a >= 3 && b >= 3 && c >= 3) {
      total += mod1;
      a -= mod1;
      b -= mod1;
      c -= mod1;
    }
    total += a / 3 + b / 3 + c / 3;
  } else {
    total += a / 3 + b / 3 + c / 3;
  }
  cout << total << endl;
  return 0;
}

```