#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int t, i, j, n, m, ans, sum = 0;
  long long int r, g, b, r1, b1, g1;
  cin >> r >> g >> b;
  r1 = r;
  g1 = g;
  b1 = b;
  sum += min(min(r, g), b);
  r -= sum;
  g -= sum;
  b -= sum;
  if (sum != 0) {
    if (r <= g && r <= b) {
      if (g % 3 == 2 && b % 3 == 2) {
        g++;
        b++;
        r++;
        sum--;
      }
    } else if (g <= r && g <= b) {
      if (r % 3 == 2 && b % 3 == 2) {
        g++;
        b++;
        r++;
        sum--;
      }
    } else {
      if (g % 3 == 2 && r % 3 == 2) {
        g++;
        b++;
        r++;
        sum--;
      }
    }
  }
  sum += r / 3 + g / 3 + b / 3;
  ans = r1 / 3 + g1 / 3 + b1 / 3;
  r1 %= 3;
  g1 %= 3;
  b1 %= 3;
  cout << max(ans + min(r1, min(b1, g1)), sum);
  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>
using namespace std;
int main() {
  long long int t, i, j, n, m, ans, sum = 0;
  long long int r, g, b, r1, b1, g1;
  cin >> r >> g >> b;
  r1 = r;
  g1 = g;
  b1 = b;
  sum += min(min(r, g), b);
  r -= sum;
  g -= sum;
  b -= sum;
  if (sum != 0) {
    if (r <= g && r <= b) {
      if (g % 3 == 2 && b % 3 == 2) {
        g++;
        b++;
        r++;
        sum--;
      }
    } else if (g <= r && g <= b) {
      if (r % 3 == 2 && b % 3 == 2) {
        g++;
        b++;
        r++;
        sum--;
      }
    } else {
      if (g % 3 == 2 && r % 3 == 2) {
        g++;
        b++;
        r++;
        sum--;
      }
    }
  }
  sum += r / 3 + g / 3 + b / 3;
  ans = r1 / 3 + g1 / 3 + b1 / 3;
  r1 %= 3;
  g1 %= 3;
  b1 %= 3;
  cout << max(ans + min(r1, min(b1, g1)), sum);
  return 0;
}

```