#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, c;
  cin >> a >> b >> c;
  long long ans1 = min(a, min(b, c));
  long long tmp = (a - ans1) / 3 + (b - ans1) / 3 + (c - ans1) / 3;
  int cnt1 = 0;
  int cnt = 0;
  if ((a - ans1) % 3 == 2) cnt1++;
  if ((b - ans1) % 3 == 2) cnt1++;
  if ((c - ans1) % 3 == 2) cnt1++;
  if ((a - ans1) % 3 == 0) cnt++;
  if ((b - ans1) % 3 == 0) cnt++;
  if ((c - ans1) % 3 == 0) cnt++;
  ans1 += tmp;
  if (cnt1 == 2 && (a != 0) && (b != 0) && (c != 0)) {
    ans1 += 1;
  }
  long long ans2 = 0;
  if (a > 3) {
    ans2 += a / 3;
    a %= 3;
  }
  if (b > 3) {
    ans2 += b / 3;
    b %= 3;
  }
  if (c > 3) {
    ans2 += c / 3;
    c %= 3;
  }
  ans2 += min(a, min(b, c));
  cout << max(ans1, ans2) << endl;
}

### Prompt

Your task is to create 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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, c;
  cin >> a >> b >> c;
  long long ans1 = min(a, min(b, c));
  long long tmp = (a - ans1) / 3 + (b - ans1) / 3 + (c - ans1) / 3;
  int cnt1 = 0;
  int cnt = 0;
  if ((a - ans1) % 3 == 2) cnt1++;
  if ((b - ans1) % 3 == 2) cnt1++;
  if ((c - ans1) % 3 == 2) cnt1++;
  if ((a - ans1) % 3 == 0) cnt++;
  if ((b - ans1) % 3 == 0) cnt++;
  if ((c - ans1) % 3 == 0) cnt++;
  ans1 += tmp;
  if (cnt1 == 2 && (a != 0) && (b != 0) && (c != 0)) {
    ans1 += 1;
  }
  long long ans2 = 0;
  if (a > 3) {
    ans2 += a / 3;
    a %= 3;
  }
  if (b > 3) {
    ans2 += b / 3;
    b %= 3;
  }
  if (c > 3) {
    ans2 += c / 3;
    c %= 3;
  }
  ans2 += min(a, min(b, c));
  cout << max(ans1, ans2) << endl;
}

```