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;
void boost() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
template <typename T>
void take(vector<T>& v) {
  for (T& val : v) cin >> val;
}
template <typename T>
void show(vector<T>& v, string end = " ", string after = "\n") {
  for (T& val : v) cout << val << end;
  cout << after;
}
void solve() {}
int main() {
  boost();
  int r, g, b;
  cin >> r >> g >> b;
  if (r == 0 || g == 0 || b == 0) {
    cout << (r / 3 + g / 3 + b / 3) << endl;
    return (0);
  }
  int ans = min(r, min(g, b));
  r -= ans;
  g -= ans;
  b -= ans;
  ans += r / 3 + g / 3 + b / 3;
  r %= 3;
  g %= 3;
  b %= 3;
  if (r == 0 && g == 2 && b == 2) ans++;
  if (g == 0 && r == 2 && b == 2) ans++;
  if (b == 0 && r == 2 && g == 2) ans++;
  cout << ans << endl;
  return (0);
}
