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;
long long int vmax(vector<long long int> v) {
  long long int max = v[0];
  long long int j;
  for (j = 0; j < v.size(); j++) {
    if (v[j] > max) {
      max = v[j];
    }
  }
  return (max);
}
long long int vmin(vector<long long int> v) {
  long long int min = v[0];
  long long int j;
  for (j = 0; j < v.size(); j++) {
    if (v[j] < min) {
      min = v[j];
    }
  }
  return (min);
}
void solve() {
  vector<long long int> v;
  long long int r, g, b;
  cin >> r >> g >> b;
  v.push_back(r);
  v.push_back(g);
  v.push_back(b);
  sort(v.begin(), v.end());
  long long int ans = 0;
  long long int mm;
  long long int x = 3;
  mm = min(v[0] + 1, x);
  for (int j = 0; j < mm; j += 1) {
    long long int a, b, c;
    a = v[0];
    b = v[1];
    c = v[2];
    a = a - j;
    b = b - j;
    c = c - j;
    long long int y;
    y = j + a / 3 + b / 3 + c / 3;
    ans = max(ans, y);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
  }
}
