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.
#include <bits/stdc++.h>
using namespace std;
bool IsPrime(long long x) {
  bool isPrime = true;
  for (long long a = 2; a * a <= x; a++) {
    if (x % a == 0) return false;
  }
  return true;
}
long long A[1000010], B[1000010], C[200002];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a, b, c, m, d;
  cin >> a >> b >> c;
  if (a % 3 == 0 && b % 3 == 0 && c % 3 == 0) {
    cout << a / 3 + b / 3 + c / 3;
    return 0;
  }
  a--;
  b--;
  c--;
  d = a / 3 + b / 3 + c / 3;
  a %= 3;
  b %= 3;
  c %= 3;
  a++;
  b++;
  c++;
  if (a == b && b == c) {
    cout << (a + b + c) / 3 + d;
    return 0;
  }
  m = min(a, b);
  m = min(c, m);
  long long mm = max(a, b);
  mm = max(c, mm);
  if (m == 2) {
    cout << d + 2;
    return 0;
  }
  if ((m == 1 && mm == 3) || m == 0) {
    cout << a / 3 + b / 3 + c / 3 + d;
    return 0;
  }
  if (m == 1 && mm == 2) cout << d + 1;
}
