#include <bits/stdc++.h>
using namespace std;
bool cmp(int xx, int yy) { return xx > yy; }
int main() {
  int r, b, g;
  while (cin >> r >> g >> b) {
    int r1 = r % 3, g1 = g % 3, b1 = b % 3;
    int ans = r / 3 + g / 3 + b / 3;
    if (!((((r) < (g) ? (r) : (g))) < (b) ? (((r) < (g) ? (r) : (g))) : (b)))
      cout << ans << endl;
    else if (((((r) > (g) ? (r) : (g))) > (b) ? (((r) > (g) ? (r) : (g)))
                                              : (b)) < 3)
      cout << ((((r) < (g) ? (r) : (g))) < (b) ? (((r) < (g) ? (r) : (g)))
                                               : (b))
           << endl;
    else {
      while (((((r1) < (g1) ? (r1) : (g1))) < (b1)
                  ? (((r1) < (g1) ? (r1) : (g1)))
                  : (b1))) {
        r1--;
        g1--;
        b1--;
        ans++;
      }
      if (r1 + b1 + g1 == 4) ans++;
      cout << ans << endl;
    }
  }
  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;
bool cmp(int xx, int yy) { return xx > yy; }
int main() {
  int r, b, g;
  while (cin >> r >> g >> b) {
    int r1 = r % 3, g1 = g % 3, b1 = b % 3;
    int ans = r / 3 + g / 3 + b / 3;
    if (!((((r) < (g) ? (r) : (g))) < (b) ? (((r) < (g) ? (r) : (g))) : (b)))
      cout << ans << endl;
    else if (((((r) > (g) ? (r) : (g))) > (b) ? (((r) > (g) ? (r) : (g)))
                                              : (b)) < 3)
      cout << ((((r) < (g) ? (r) : (g))) < (b) ? (((r) < (g) ? (r) : (g)))
                                               : (b))
           << endl;
    else {
      while (((((r1) < (g1) ? (r1) : (g1))) < (b1)
                  ? (((r1) < (g1) ? (r1) : (g1)))
                  : (b1))) {
        r1--;
        g1--;
        b1--;
        ans++;
      }
      if (r1 + b1 + g1 == 4) ans++;
      cout << ans << endl;
    }
  }
  return 0;
}

```