#include <bits/stdc++.h>
int r, g, b, s, min, a[4], i;
int main() {
  scanf("%d %d %d", &r, &g, &b);
  a[0] = r;
  a[1] = g;
  a[2] = b;
  min = 1000000001;
  for (i = 0; i < 3; i++) {
    if (min > a[i]) {
      min = a[i];
    }
  }
  r = r - min;
  g = g - min;
  b = b - min;
  s = r / 3 + g / 3 + b / 3 + min;
  if (r % 3 + g % 3 + b % 3 == 4 && min != 0) {
    s++;
  }
  printf("%d", s);
  return 0;
}

### Prompt

Generate 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>
int r, g, b, s, min, a[4], i;
int main() {
  scanf("%d %d %d", &r, &g, &b);
  a[0] = r;
  a[1] = g;
  a[2] = b;
  min = 1000000001;
  for (i = 0; i < 3; i++) {
    if (min > a[i]) {
      min = a[i];
    }
  }
  r = r - min;
  g = g - min;
  b = b - min;
  s = r / 3 + g / 3 + b / 3 + min;
  if (r % 3 + g % 3 + b % 3 == 4 && min != 0) {
    s++;
  }
  printf("%d", s);
  return 0;
}

```