#include <bits/stdc++.h>
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
const int MOD = 1e12 + 7;
const int MAX = 1e6 + 2;
const int INF = 0x3f3f3f3f;
bool IsPrime(long long x) {
  if (x < 2) {
    return false;
  }
  for (long long p = 2; p * p <= x; ++p)
    if (x % p == 0) return false;
  return true;
}
using namespace std;
int main() {
  long long r, g, b, cnt = 0;
  cin >> r >> g >> b;
  for (long long i = 0; i < 3; i++) {
    if (i > r || i > b || i > g) break;
    cnt = max(cnt, i + (r - i) / 3 + (g - i) / 3 + (b - i) / 3);
  }
  cout << cnt << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
const int MOD = 1e12 + 7;
const int MAX = 1e6 + 2;
const int INF = 0x3f3f3f3f;
bool IsPrime(long long x) {
  if (x < 2) {
    return false;
  }
  for (long long p = 2; p * p <= x; ++p)
    if (x % p == 0) return false;
  return true;
}
using namespace std;
int main() {
  long long r, g, b, cnt = 0;
  cin >> r >> g >> b;
  for (long long i = 0; i < 3; i++) {
    if (i > r || i > b || i > g) break;
    cnt = max(cnt, i + (r - i) / 3 + (g - i) / 3 + (b - i) / 3);
  }
  cout << cnt << endl;
  return 0;
}

```