#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &S, const vector<T> &v) {
  S << "(";
  for (int i = 0; i < v.size(); i++) {
    if (i) S << ", ";
    S << v[i];
  }
  S << ")";
  return S;
}
template <class T1, class T2>
ostream &operator<<(ostream &S, const pair<T1, T2> &p) {
  S << "(" << p.first << ", " << p.second << ")";
  return S;
}
template <class T>
T abs(T num) {
  return num > 0 ? num : -num;
}
template <class T>
T sqr(T num) {
  return num * num;
}
int main() {
  int r, g, b;
  cin >> r >> g >> b;
  int ans = ~0x7FFFFFFF;
  int m = min(100, min(r, min(g, b)));
  for (int i = 0; i <= m; i++) {
    int cur = i;
    cur += (r - i) / 3;
    cur += (g - i) / 3;
    cur += (b - i) / 3;
    ans = max(ans, cur);
  }
  printf("%d\n", ans);
  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;
template <class T>
ostream &operator<<(ostream &S, const vector<T> &v) {
  S << "(";
  for (int i = 0; i < v.size(); i++) {
    if (i) S << ", ";
    S << v[i];
  }
  S << ")";
  return S;
}
template <class T1, class T2>
ostream &operator<<(ostream &S, const pair<T1, T2> &p) {
  S << "(" << p.first << ", " << p.second << ")";
  return S;
}
template <class T>
T abs(T num) {
  return num > 0 ? num : -num;
}
template <class T>
T sqr(T num) {
  return num * num;
}
int main() {
  int r, g, b;
  cin >> r >> g >> b;
  int ans = ~0x7FFFFFFF;
  int m = min(100, min(r, min(g, b)));
  for (int i = 0; i <= m; i++) {
    int cur = i;
    cur += (r - i) / 3;
    cur += (g - i) / 3;
    cur += (b - i) / 3;
    ans = max(ans, cur);
  }
  printf("%d\n", ans);
  return 0;
}

```