#include <bits/stdc++.h>
using namespace std;
long long inv(long long a, long long b) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
void solve() {
  long long r, g, b;
  cin >> r >> g >> b;
  if (g < r) swap(g, r);
  if (b < r) swap(b, r);
  if (g > b) swap(g, b);
  long long x = g - r;
  long long cnt = 0, mini = r;
  if (x % 3 == 0)
    cnt += (x / 3);
  else if (x % 3 == 2) {
    cnt += (x / 3);
    if (r > 0) {
      cnt++;
      mini = r - 1;
    }
  } else if (x > 0 and x % 3 == 1) {
    x--;
    cnt += (x / 3);
    mini = min(mini, r + 1);
  }
  x = b - r;
  if (x % 3 == 0)
    cnt += (x / 3);
  else if (x % 3 == 2) {
    cnt += (x / 3);
    if (r > 0) {
      cnt++;
      mini = r - 1;
    }
  } else if (x > 0 and x % 3 == 1) {
    x--;
    cnt += (x / 3);
    mini = min(mini, r + 1);
  }
  cnt += mini;
  cout << cnt;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}

### 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;
long long inv(long long a, long long b) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
void solve() {
  long long r, g, b;
  cin >> r >> g >> b;
  if (g < r) swap(g, r);
  if (b < r) swap(b, r);
  if (g > b) swap(g, b);
  long long x = g - r;
  long long cnt = 0, mini = r;
  if (x % 3 == 0)
    cnt += (x / 3);
  else if (x % 3 == 2) {
    cnt += (x / 3);
    if (r > 0) {
      cnt++;
      mini = r - 1;
    }
  } else if (x > 0 and x % 3 == 1) {
    x--;
    cnt += (x / 3);
    mini = min(mini, r + 1);
  }
  x = b - r;
  if (x % 3 == 0)
    cnt += (x / 3);
  else if (x % 3 == 2) {
    cnt += (x / 3);
    if (r > 0) {
      cnt++;
      mini = r - 1;
    }
  } else if (x > 0 and x % 3 == 1) {
    x--;
    cnt += (x / 3);
    mini = min(mini, r + 1);
  }
  cnt += mini;
  cout << cnt;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```