#include <bits/stdc++.h>
using namespace std;
vector<bool> sieve(long long n) {
  vector<bool> prime(n + 1, true);
  prime[0] = prime[1] = false;
  for (long long i = 2; i <= n; ++i) {
    if (prime[i] && (i * i) <= n)
      for (long long j = i * i; j <= n; j += i) prime[j] = false;
  }
  return prime;
}
long long power(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long r, g, b;
  cin >> r >> g >> b;
  if (r * g * b == 0) {
    cout << r / 3 + b / 3 + g / 3;
    return 0;
  }
  long long val1 = r / 3 + b / 3 + g / 3;
  long long val2 = 1 + (r - 1) / 3 + (b - 1) / 3 + (g - 1) / 3;
  long long val3 = 2 + (r - 2) / 3 + (b - 2) / 3 + (g - 2) / 3;
  cout << max({val1, val2, val3});
  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;
vector<bool> sieve(long long n) {
  vector<bool> prime(n + 1, true);
  prime[0] = prime[1] = false;
  for (long long i = 2; i <= n; ++i) {
    if (prime[i] && (i * i) <= n)
      for (long long j = i * i; j <= n; j += i) prime[j] = false;
  }
  return prime;
}
long long power(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long r, g, b;
  cin >> r >> g >> b;
  if (r * g * b == 0) {
    cout << r / 3 + b / 3 + g / 3;
    return 0;
  }
  long long val1 = r / 3 + b / 3 + g / 3;
  long long val2 = 1 + (r - 1) / 3 + (b - 1) / 3 + (g - 1) / 3;
  long long val3 = 2 + (r - 2) / 3 + (b - 2) / 3 + (g - 2) / 3;
  cout << max({val1, val2, val3});
  return 0;
}

```