#include <bits/stdc++.h>
void change(int *ans, int *cnt) {
  if ((*cnt) % 3 == 0) {
    if ((*cnt) != 0) {
      (*ans) += ((*cnt) / 3 - 1);
      (*cnt) = 3;
    }
  } else {
    (*ans) += ((*cnt) / 3);
    (*cnt) %= 3;
  }
}
void solve(int r, int g, int b) {
  int ans, m;
  ans = 0;
  change(&ans, &r);
  change(&ans, &g);
  change(&ans, &b);
  m = ((r) < (g) ? (r) : (g));
  m = ((m) < (b) ? (m) : (b));
  m = ((m) > (r / 3 + g / 3 + b / 3) ? (m) : (r / 3 + g / 3 + b / 3));
  ans += m;
  printf("%d\n", ans);
}
int main() {
  int r, g, b;
  while (scanf("%d%d%d", &r, &g, &b) == 3) {
    solve(r, g, b);
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
void change(int *ans, int *cnt) {
  if ((*cnt) % 3 == 0) {
    if ((*cnt) != 0) {
      (*ans) += ((*cnt) / 3 - 1);
      (*cnt) = 3;
    }
  } else {
    (*ans) += ((*cnt) / 3);
    (*cnt) %= 3;
  }
}
void solve(int r, int g, int b) {
  int ans, m;
  ans = 0;
  change(&ans, &r);
  change(&ans, &g);
  change(&ans, &b);
  m = ((r) < (g) ? (r) : (g));
  m = ((m) < (b) ? (m) : (b));
  m = ((m) > (r / 3 + g / 3 + b / 3) ? (m) : (r / 3 + g / 3 + b / 3));
  ans += m;
  printf("%d\n", ans);
}
int main() {
  int r, g, b;
  while (scanf("%d%d%d", &r, &g, &b) == 3) {
    solve(r, g, b);
  }
  return 0;
}

```