#include <bits/stdc++.h>
long long int min(long long int x, long long int y, long long int z) {
  if (x > y) x = y;
  if (x > z) return z;
  return x;
}
long long int max(long long int x, long long int y) {
  if (x > y)
    return x;
  else
    return y;
}
int main() {
  long long int x, y, z, ans = 0;
  scanf("%lld%lld%lld", &x, &y, &z);
  long long int t = min(x, y, z);
  ans += t;
  x -= ans;
  y -= ans;
  z -= ans;
  ans += x / 3 + y / 3 + z / 3;
  x %= 3;
  y %= 3;
  z %= 3;
  if (t) {
    t--;
    x++;
    y++;
    z++;
    ans = max(ans, ans + x / 3 + y / 3 + z / 3 - 1);
  }
  printf("%lld", ans);
  return 0;
}

### 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>
long long int min(long long int x, long long int y, long long int z) {
  if (x > y) x = y;
  if (x > z) return z;
  return x;
}
long long int max(long long int x, long long int y) {
  if (x > y)
    return x;
  else
    return y;
}
int main() {
  long long int x, y, z, ans = 0;
  scanf("%lld%lld%lld", &x, &y, &z);
  long long int t = min(x, y, z);
  ans += t;
  x -= ans;
  y -= ans;
  z -= ans;
  ans += x / 3 + y / 3 + z / 3;
  x %= 3;
  y %= 3;
  z %= 3;
  if (t) {
    t--;
    x++;
    y++;
    z++;
    ans = max(ans, ans + x / 3 + y / 3 + z / 3 - 1);
  }
  printf("%lld", ans);
  return 0;
}

```