#include <bits/stdc++.h>
using namespace std;
int tot;
int maxn;
void dfs(int r, int g, int b, int now) {
  if (now > maxn) maxn = now;
  if (r - 3 >= 0) dfs(r - 3, g, b, now + 1);
  if (g - 3 >= 0) dfs(r, g - 3, b, now + 1);
  if (b - 3 >= 0) dfs(r, g, b - 3, now + 1);
  if (r >= 1 && g >= 1 && b >= 1) dfs(r - 1, g - 1, b - 1, now + 1);
}
int main() {
  int r, g, b;
  while (scanf("%d%d%d", &r, &g, &b) != EOF) {
    maxn = 0;
    tot = 0;
    if (r > 3) {
      tot += (r - 3) / 3;
      r = ((r - 3) % 3) + 3;
    }
    if (g > 3) {
      tot += (g - 3) / 3;
      g = ((g - 3) % 3) + 3;
    }
    if (b > 3) {
      tot += (b - 3) / 3;
      b = ((b - 3) % 3) + 3;
    }
    dfs(r, g, b, 0);
    printf("%d\n", tot + maxn);
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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;
int tot;
int maxn;
void dfs(int r, int g, int b, int now) {
  if (now > maxn) maxn = now;
  if (r - 3 >= 0) dfs(r - 3, g, b, now + 1);
  if (g - 3 >= 0) dfs(r, g - 3, b, now + 1);
  if (b - 3 >= 0) dfs(r, g, b - 3, now + 1);
  if (r >= 1 && g >= 1 && b >= 1) dfs(r - 1, g - 1, b - 1, now + 1);
}
int main() {
  int r, g, b;
  while (scanf("%d%d%d", &r, &g, &b) != EOF) {
    maxn = 0;
    tot = 0;
    if (r > 3) {
      tot += (r - 3) / 3;
      r = ((r - 3) % 3) + 3;
    }
    if (g > 3) {
      tot += (g - 3) / 3;
      g = ((g - 3) % 3) + 3;
    }
    if (b > 3) {
      tot += (b - 3) / 3;
      b = ((b - 3) % 3) + 3;
    }
    dfs(r, g, b, 0);
    printf("%d\n", tot + maxn);
  }
  return 0;
}

```