#include <bits/stdc++.h>
bool cmp(const int &a, const int &b) { return (a < b); }
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
int main() {
  int a[3], b[3], c[3];
  scanf("%d%d%d", &a[0], &a[1], &a[2]);
  std::sort(a, a + 3, cmp);
  for (int i = 0; i < 3; i++) {
    b[i] = a[i];
    c[i] = a[i];
  }
  int ans1 = 0, ans2 = 0, ans3 = 0;
  if (a[0] >= 2) {
    for (int i = 0; i < 3; i++) {
      b[i] -= 1;
      c[i] -= 2;
    }
    ans1 += 1;
    ans2 += 2;
  }
  for (int i = 0; i < 3; i++) {
    ans1 += b[i] / 3;
    ans2 += c[i] / 3;
    ans3 += a[i] / 3;
  }
  printf("%d\n", max(max(ans1, ans2), ans3));
  return 0;
}

### Prompt

Please create a solution in CPP to 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>
bool cmp(const int &a, const int &b) { return (a < b); }
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
int main() {
  int a[3], b[3], c[3];
  scanf("%d%d%d", &a[0], &a[1], &a[2]);
  std::sort(a, a + 3, cmp);
  for (int i = 0; i < 3; i++) {
    b[i] = a[i];
    c[i] = a[i];
  }
  int ans1 = 0, ans2 = 0, ans3 = 0;
  if (a[0] >= 2) {
    for (int i = 0; i < 3; i++) {
      b[i] -= 1;
      c[i] -= 2;
    }
    ans1 += 1;
    ans2 += 2;
  }
  for (int i = 0; i < 3; i++) {
    ans1 += b[i] / 3;
    ans2 += c[i] / 3;
    ans3 += a[i] / 3;
  }
  printf("%d\n", max(max(ans1, ans2), ans3));
  return 0;
}

```