#include <bits/stdc++.h>
using namespace std;
int main() {
  long long r, b, g, s, ans1 = 0, ans2 = 0, R, B, G, loop = 0, ans = 0, i, j, k;
  cin >> r >> g >> b;
  R = r;
  B = b;
  G = g;
  long long a[3] = {r, g, b};
  sort(a, a + 3);
  ans1 = a[0];
  for (i = 0; i < 4; ++i) {
    for (j = 0; j < 4; ++j) {
      for (k = 0; k < 4; ++k) {
        r = R;
        g = G;
        b = B;
        r -= i;
        g -= j;
        b -= k;
        long long a[3] = {r, g, b};
        sort(a, a + 3);
        if (a[0] < 0) continue;
        ans1 = a[0];
        r -= a[0];
        b -= a[0];
        g -= a[0];
        ans1 += r / 3;
        ans1 += b / 3;
        ans1 += g / 3;
        ans = max(ans, ans1);
      }
    }
  }
  ans2 = R / 3 + B / 3 + G / 3;
  R %= 3;
  G %= 3;
  B %= 3;
  a[0] = R;
  a[1] = B;
  a[2] = G;
  sort(a, a + 3);
  ans2 += a[0];
  cout << max(ans, ans2);
  return 0;
}

### Prompt

Generate a cpp solution 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>
using namespace std;
int main() {
  long long r, b, g, s, ans1 = 0, ans2 = 0, R, B, G, loop = 0, ans = 0, i, j, k;
  cin >> r >> g >> b;
  R = r;
  B = b;
  G = g;
  long long a[3] = {r, g, b};
  sort(a, a + 3);
  ans1 = a[0];
  for (i = 0; i < 4; ++i) {
    for (j = 0; j < 4; ++j) {
      for (k = 0; k < 4; ++k) {
        r = R;
        g = G;
        b = B;
        r -= i;
        g -= j;
        b -= k;
        long long a[3] = {r, g, b};
        sort(a, a + 3);
        if (a[0] < 0) continue;
        ans1 = a[0];
        r -= a[0];
        b -= a[0];
        g -= a[0];
        ans1 += r / 3;
        ans1 += b / 3;
        ans1 += g / 3;
        ans = max(ans, ans1);
      }
    }
  }
  ans2 = R / 3 + B / 3 + G / 3;
  R %= 3;
  G %= 3;
  B %= 3;
  a[0] = R;
  a[1] = B;
  a[2] = G;
  sort(a, a + 3);
  ans2 += a[0];
  cout << max(ans, ans2);
  return 0;
}

```