#include <bits/stdc++.h>
using namespace std;
long long int r, g, b;
void solve() {
  vector<long long int> cnt(3, 0);
  long long int mn = min(r, min(g, b));
  cnt[r % 3]++;
  cnt[b % 3]++;
  cnt[g % 3]++;
  long long int ans = (r / 3) + (b / 3) + (g / 3);
  if (cnt[0] == cnt[1] && cnt[1] == cnt[2]) {
    cout << ans;
  } else if (cnt[0] >= 2) {
    cout << ans;
  } else if (cnt[1] >= 2) {
    r--;
    b--;
    g--;
    long long int gg = (r / 3) + (b / 3) + (g / 3);
    if (mn >= 1)
      cout << gg + 1;
    else
      cout << gg;
  } else {
    r -= 2;
    b -= 2;
    g -= 2;
    long long int gg = (r / 3) + (b / 3) + (g / 3);
    if (mn >= 2)
      cout << gg + 2;
    else
      cout << gg + mn;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> r >> g >> b;
  solve();
  return 0;
}

### Prompt

Please formulate 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;
long long int r, g, b;
void solve() {
  vector<long long int> cnt(3, 0);
  long long int mn = min(r, min(g, b));
  cnt[r % 3]++;
  cnt[b % 3]++;
  cnt[g % 3]++;
  long long int ans = (r / 3) + (b / 3) + (g / 3);
  if (cnt[0] == cnt[1] && cnt[1] == cnt[2]) {
    cout << ans;
  } else if (cnt[0] >= 2) {
    cout << ans;
  } else if (cnt[1] >= 2) {
    r--;
    b--;
    g--;
    long long int gg = (r / 3) + (b / 3) + (g / 3);
    if (mn >= 1)
      cout << gg + 1;
    else
      cout << gg;
  } else {
    r -= 2;
    b -= 2;
    g -= 2;
    long long int gg = (r / 3) + (b / 3) + (g / 3);
    if (mn >= 2)
      cout << gg + 2;
    else
      cout << gg + mn;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> r >> g >> b;
  solve();
  return 0;
}

```