#include <bits/stdc++.h>
using namespace std;
int ary[4];
int sum;
int main() {
  while (~scanf("%d%d%d", &ary[1], &ary[2], &ary[3])) {
    sum = 0;
    for (int i = 1; i <= 3; i++) {
      if (ary[i] != 0) {
        if (ary[i] % 3 == 0) {
          sum += ary[i] / 3;
          sum--;
          ary[i] = 3;
        } else {
          sum += ary[i] / 3;
          ary[i] %= 3;
        }
      }
    }
    int count = 0;
    for (int i = 1; i <= 3; i++) {
      if (ary[i] == 3) count++;
    }
    if (count == 3)
      printf("%d\n", sum + 3);
    else if (count == 2) {
      printf("%d\n", sum + 2);
    } else if (count == 1) {
      int num = 0;
      for (int i = 1; i <= 3; i++) {
        if (ary[i] == 2) num++;
      }
      if (num == 2)
        printf("%d\n", sum + 2);
      else
        printf("%d\n", sum + 1);
    } else {
      int min = 10;
      for (int i = 1; i <= 3; i++) {
        if (min > ary[i]) min = ary[i];
      }
      printf("%d\n", sum + min);
    }
  }
}

### Prompt

In Cpp, your task is to solve 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 ary[4];
int sum;
int main() {
  while (~scanf("%d%d%d", &ary[1], &ary[2], &ary[3])) {
    sum = 0;
    for (int i = 1; i <= 3; i++) {
      if (ary[i] != 0) {
        if (ary[i] % 3 == 0) {
          sum += ary[i] / 3;
          sum--;
          ary[i] = 3;
        } else {
          sum += ary[i] / 3;
          ary[i] %= 3;
        }
      }
    }
    int count = 0;
    for (int i = 1; i <= 3; i++) {
      if (ary[i] == 3) count++;
    }
    if (count == 3)
      printf("%d\n", sum + 3);
    else if (count == 2) {
      printf("%d\n", sum + 2);
    } else if (count == 1) {
      int num = 0;
      for (int i = 1; i <= 3; i++) {
        if (ary[i] == 2) num++;
      }
      if (num == 2)
        printf("%d\n", sum + 2);
      else
        printf("%d\n", sum + 1);
    } else {
      int min = 10;
      for (int i = 1; i <= 3; i++) {
        if (min > ary[i]) min = ary[i];
      }
      printf("%d\n", sum + min);
    }
  }
}

```