#include <bits/stdc++.h>
using namespace std;
int main() {
  int a[3];
  long long ctr = 0;
  for (int i = 0; i < 3; i++) {
    cin >> a[i];
  }
  long long zeroMixed = 0;
  for (int i = 0; i < 3; i++) {
    zeroMixed += (a[i] / 3);
  }
  sort(a, a + 3);
  long long oneMixed = -1;
  if (a[0] >= 1) {
    oneMixed = 1;
    for (int i = 0; i < 3; i++) {
      oneMixed += ((a[i] - 1) / 3);
    }
  }
  long long twoMixed = -1;
  if (a[0] >= 2) {
    twoMixed = 2;
    for (int i = 0; i < 3; i++) {
      twoMixed += ((a[i] - 2) / 3);
    }
  }
  cout << max(twoMixed, max(oneMixed, zeroMixed));
  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>
using namespace std;
int main() {
  int a[3];
  long long ctr = 0;
  for (int i = 0; i < 3; i++) {
    cin >> a[i];
  }
  long long zeroMixed = 0;
  for (int i = 0; i < 3; i++) {
    zeroMixed += (a[i] / 3);
  }
  sort(a, a + 3);
  long long oneMixed = -1;
  if (a[0] >= 1) {
    oneMixed = 1;
    for (int i = 0; i < 3; i++) {
      oneMixed += ((a[i] - 1) / 3);
    }
  }
  long long twoMixed = -1;
  if (a[0] >= 2) {
    twoMixed = 2;
    for (int i = 0; i < 3; i++) {
      twoMixed += ((a[i] - 2) / 3);
    }
  }
  cout << max(twoMixed, max(oneMixed, zeroMixed));
  return 0;
}

```