#include <bits/stdc++.h>
int main() {
  long long a[3], i, k, ans = 0, min;
  int max = 0;
  scanf("%llu %llu %llu", &a[0], &a[1], &a[2]);
  for (i = 0; i < 3; i++) {
    k = a[i] / 3;
    if (a[i] % 3 == 0 && a[i] != 0) {
      k--;
    }
    ans += k;
    a[i] -= k * 3;
  }
  min = a[0];
  max = a[0] / 3 + a[1] / 3 + a[2] / 3;
  for (i = 1; i < 3; i++) {
    if (a[i] < min) {
      min = a[i];
    }
  }
  k = 0;
  if (max > min) {
    k = max;
  } else {
    k = min;
  }
  printf("%llu", ans + k);
  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>
int main() {
  long long a[3], i, k, ans = 0, min;
  int max = 0;
  scanf("%llu %llu %llu", &a[0], &a[1], &a[2]);
  for (i = 0; i < 3; i++) {
    k = a[i] / 3;
    if (a[i] % 3 == 0 && a[i] != 0) {
      k--;
    }
    ans += k;
    a[i] -= k * 3;
  }
  min = a[0];
  max = a[0] / 3 + a[1] / 3 + a[2] / 3;
  for (i = 1; i < 3; i++) {
    if (a[i] < min) {
      min = a[i];
    }
  }
  k = 0;
  if (max > min) {
    k = max;
  } else {
    k = min;
  }
  printf("%llu", ans + k);
  return 0;
}

```