#include <bits/stdc++.h>
using namespace std;
string s1[1001];
long long r, g, b, t1, t2, t3, ans;
map<string, int> mp, tp;
int main() {
  cin >> r >> g >> b;
  if (r != 0 && g != 0 && b != 0) {
    t1 = r % 3;
    t2 = g % 3;
    t3 = b % 3;
  }
  if (t1 != 0 && t2 != 0 && t3 == 0)
    t3 = 3;
  else if (t1 != 0 && t3 != 0 && t2 == 0)
    t2 = 3;
  else if (t2 != 0 && t3 != 0 && t1 == 0)
    t1 = 3;
  while (t1 > 0 && t2 > 0 && t3 > 0) {
    ans++;
    t1--;
    t2--;
    t3--;
  }
  r -= t1;
  g -= t2;
  b -= t3;
  ans += (r / 3 + g / 3 + b / 3);
  cout << ans;
}

### Prompt

Your challenge is to write 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;
string s1[1001];
long long r, g, b, t1, t2, t3, ans;
map<string, int> mp, tp;
int main() {
  cin >> r >> g >> b;
  if (r != 0 && g != 0 && b != 0) {
    t1 = r % 3;
    t2 = g % 3;
    t3 = b % 3;
  }
  if (t1 != 0 && t2 != 0 && t3 == 0)
    t3 = 3;
  else if (t1 != 0 && t3 != 0 && t2 == 0)
    t2 = 3;
  else if (t2 != 0 && t3 != 0 && t1 == 0)
    t1 = 3;
  while (t1 > 0 && t2 > 0 && t3 > 0) {
    ans++;
    t1--;
    t2--;
    t3--;
  }
  r -= t1;
  g -= t2;
  b -= t3;
  ans += (r / 3 + g / 3 + b / 3);
  cout << ans;
}

```