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;
int main() {
  int r, g, b;
  int rb = 0, gb = 0, bb = 0, mb = 0, mn;
  cin >> r >> g >> b;
  int ch1, ch2, ch3, r1, g1, b1;
  if (r < 1 && g < 1 && b < 1) {
    cout << "0" << endl;
  } else {
    if (r < 3 && g < 3 && b < 3) {
      mb = min(r, min(g, b));
      cout << mb << endl;
    } else {
      mn = min(r, min(b, g));
      ch1 = INT_MIN;
      if (mn > 2) {
        for (int i = 0; i <= 2; i++) {
          r1 = r - i;
          g1 = g - i;
          b1 = b - i;
          rb = r1 / 3;
          gb = g1 / 3;
          bb = b1 / 3;
          if (ch1 < (i + rb + gb + bb)) {
            ch1 = (i + rb + gb + bb);
          }
        }
      } else {
        for (int i = 0; i <= mn; i++) {
          r1 = r - i;
          g1 = g - i;
          b1 = b - i;
          rb = r1 / 3;
          gb = g1 / 3;
          bb = b1 / 3;
          if (ch1 < (i + rb + gb + bb)) {
            ch1 = (i + rb + gb + bb);
          }
        }
      }
      cout << ch1 << endl;
    }
  }
  return 0;
}
