#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int r;
  cin >> r;
  ;
  int b;
  cin >> b;
  ;
  int g;
  cin >> g;
  ;
  bool rf, bf, gf;
  rf = bf = gf = 0;
  int ans = r / 3;
  ans += b / 3;
  ans += g / 3;
  if (r % 3 == 2) rf++;
  if (b % 3 == 2) bf++;
  if (g % 3 == 2) gf++;
  if (rf && bf && gf) {
    ans += 2;
  } else if (rf && bf) {
    if (g > 0 && g % 3 == 0)
      ans += 1;
    else if (g % 3 >= 1)
      ans += 1;
  } else if (rf && gf) {
    if (b > 0 && b % 3 == 0)
      ans += 1;
    else if (b % 3 >= 1)
      ans += 1;
  } else if (bf && gf) {
    if (r > 0 && r % 3 == 0)
      ans += 1;
    else if (r % 3 >= 1)
      ans += 1;
  } else if (r % 3 > 0 && g % 3 > 0 && b % 3 > 0)
    ans += 1;
  cout << ans;
}

### Prompt

Create a solution in CPP for 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(false);
  cin.tie(0);
  cout.tie(0);
  int r;
  cin >> r;
  ;
  int b;
  cin >> b;
  ;
  int g;
  cin >> g;
  ;
  bool rf, bf, gf;
  rf = bf = gf = 0;
  int ans = r / 3;
  ans += b / 3;
  ans += g / 3;
  if (r % 3 == 2) rf++;
  if (b % 3 == 2) bf++;
  if (g % 3 == 2) gf++;
  if (rf && bf && gf) {
    ans += 2;
  } else if (rf && bf) {
    if (g > 0 && g % 3 == 0)
      ans += 1;
    else if (g % 3 >= 1)
      ans += 1;
  } else if (rf && gf) {
    if (b > 0 && b % 3 == 0)
      ans += 1;
    else if (b % 3 >= 1)
      ans += 1;
  } else if (bf && gf) {
    if (r > 0 && r % 3 == 0)
      ans += 1;
    else if (r % 3 >= 1)
      ans += 1;
  } else if (r % 3 > 0 && g % 3 > 0 && b % 3 > 0)
    ans += 1;
  cout << ans;
}

```