#include <bits/stdc++.h>
using namespace std;
int main() {
  int min(int, int, int);
  int r, g, b, n;
  cin >> r >> g >> b;
  int rn = r / 3;
  int gn = g / 3;
  int bn = b / 3;
  n = rn + gn + bn;
  r %= 3;
  g %= 3;
  b %= 3;
  n += min(r, g, b);
  if (r == 0 && g == 2 && b == 2) {
    if (rn > 0) n++;
  }
  if (g == 0 && r == 2 && b == 2) {
    if (gn > 0) n++;
  }
  if (b == 0 && r == 2 && g == 2) {
    if (bn > 0) n++;
  }
  cout << n << endl;
  return 0;
}
int min(int a, int b, int c) {
  if (a < b) {
    if (a < c)
      return a;
    else
      return c;
  } else {
    if (b < c)
      return b;
    else
      return c;
  }
}

### Prompt

In Cpp, your task is to solve 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;
int main() {
  int min(int, int, int);
  int r, g, b, n;
  cin >> r >> g >> b;
  int rn = r / 3;
  int gn = g / 3;
  int bn = b / 3;
  n = rn + gn + bn;
  r %= 3;
  g %= 3;
  b %= 3;
  n += min(r, g, b);
  if (r == 0 && g == 2 && b == 2) {
    if (rn > 0) n++;
  }
  if (g == 0 && r == 2 && b == 2) {
    if (gn > 0) n++;
  }
  if (b == 0 && r == 2 && g == 2) {
    if (bn > 0) n++;
  }
  cout << n << endl;
  return 0;
}
int min(int a, int b, int c) {
  if (a < b) {
    if (a < c)
      return a;
    else
      return c;
  } else {
    if (b < c)
      return b;
    else
      return c;
  }
}

```