#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
int r, b, g;
int main() {
  while (~scanf("%d%d%d", &r, &g, &b)) {
    int ans = 0;
    int Min = min(r, min(g, b));
    if (Min >= 2) {
      int t1 = r / 3 + g / 3 + b / 3;
      int t2 = 1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3;
      int t3 = 2 + (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3;
      ans = max(t1, max(t2, t3));
    } else if (Min == 1) {
      int t1 = r / 3 + g / 3 + b / 3;
      int t2 = 1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3;
      ans = max(t1, t2);
    } else {
      ans = r / 3 + g / 3 + b / 3;
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
int r, b, g;
int main() {
  while (~scanf("%d%d%d", &r, &g, &b)) {
    int ans = 0;
    int Min = min(r, min(g, b));
    if (Min >= 2) {
      int t1 = r / 3 + g / 3 + b / 3;
      int t2 = 1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3;
      int t3 = 2 + (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3;
      ans = max(t1, max(t2, t3));
    } else if (Min == 1) {
      int t1 = r / 3 + g / 3 + b / 3;
      int t2 = 1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3;
      ans = max(t1, t2);
    } else {
      ans = r / 3 + g / 3 + b / 3;
    }
    printf("%d\n", ans);
  }
  return 0;
}

```