#include <bits/stdc++.h>
using namespace std;
template <class T, class L>
bool smax(T& x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
bool smin(T& x, L y) {
  return x > y ? (x = y, 1) : 0;
}
const int maxn = 1e2 + 17;
int a, b, c;
int get(int a, int b, int c) { return a / 3 + b / 3 + c / 3; }
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> a >> b >> c;
  int ans = get(a, b, c);
  if (a && b && c) smax(ans, get(a - 1, b - 1, c - 1) + 1);
  if (a > 1 && b > 1 && c > 1) smax(ans, get(a - 2, b - 2, c - 2) + 2);
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
template <class T, class L>
bool smax(T& x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
bool smin(T& x, L y) {
  return x > y ? (x = y, 1) : 0;
}
const int maxn = 1e2 + 17;
int a, b, c;
int get(int a, int b, int c) { return a / 3 + b / 3 + c / 3; }
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> a >> b >> c;
  int ans = get(a, b, c);
  if (a && b && c) smax(ans, get(a - 1, b - 1, c - 1) + 1);
  if (a > 1 && b > 1 && c > 1) smax(ans, get(a - 2, b - 2, c - 2) + 2);
  cout << ans << '\n';
  return 0;
}

```