#include <bits/stdc++.h>
#pragma optimize gcc("O3")
using namespace std;
long long gcd(long long a, long long b, long long* x, long long* y);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long r, g, b;
  cin >> r >> g >> b;
  long long sum = r / 3 + g / 3 + b / 3;
  long long tr = r, tg = g, tb = b;
  r %= 3;
  g %= 3;
  b %= 3;
  if (r == g && g == b)
    sum += r;
  else if (r + g + b > 3 && r != 0 && g != 0 && b != 0) {
    sum += 1;
  } else if (r + g + b > 3) {
    if (r == 0 && tr >= 3) {
      sum += 1;
    } else if (g == 0 && tg >= 3) {
      sum += 1;
    } else if (b == 0 && tb >= 3) {
      sum += 1;
    }
  }
  cout << sum << endl;
  return 0;
}
long long gcd(long long a, long long b, long long* x, long long* y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  long long x1, y1;
  long long g = gcd(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return g;
}

### 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>
#pragma optimize gcc("O3")
using namespace std;
long long gcd(long long a, long long b, long long* x, long long* y);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long r, g, b;
  cin >> r >> g >> b;
  long long sum = r / 3 + g / 3 + b / 3;
  long long tr = r, tg = g, tb = b;
  r %= 3;
  g %= 3;
  b %= 3;
  if (r == g && g == b)
    sum += r;
  else if (r + g + b > 3 && r != 0 && g != 0 && b != 0) {
    sum += 1;
  } else if (r + g + b > 3) {
    if (r == 0 && tr >= 3) {
      sum += 1;
    } else if (g == 0 && tg >= 3) {
      sum += 1;
    } else if (b == 0 && tb >= 3) {
      sum += 1;
    }
  }
  cout << sum << endl;
  return 0;
}
long long gcd(long long a, long long b, long long* x, long long* y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  long long x1, y1;
  long long g = gcd(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return g;
}

```