#include <bits/stdc++.h>
using namespace std;
string s, t;
int n;
int a[200002];
bool subseq(string s, int n, int m, int ind) {
  int i = 0;
  int j = 0;
  for (int i = 0; i < ind; i++) s[a[i] - 1] = '.';
  while (i < n && j < m) {
    if (s[i] == t[j]) j++;
    i++;
  }
  if (j < m) return false;
  return true;
}
int main() {
  {
    long long r, g, b, ans = 0;
    int a[3];
    cin >> r >> g >> b;
    a[0] = r, a[1] = g, a[2] = b;
    sort(a, a + 3);
    r = a[2], g = a[1], b = a[0];
    ans += b;
    r -= b;
    g -= b;
    ans += r / 3 + g / 3;
    if (r % 3 == 2 && g % 3 == 2 && b > 0) ans++;
    cout << ans;
  }
}

### Prompt

Your task is to create 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;
string s, t;
int n;
int a[200002];
bool subseq(string s, int n, int m, int ind) {
  int i = 0;
  int j = 0;
  for (int i = 0; i < ind; i++) s[a[i] - 1] = '.';
  while (i < n && j < m) {
    if (s[i] == t[j]) j++;
    i++;
  }
  if (j < m) return false;
  return true;
}
int main() {
  {
    long long r, g, b, ans = 0;
    int a[3];
    cin >> r >> g >> b;
    a[0] = r, a[1] = g, a[2] = b;
    sort(a, a + 3);
    r = a[2], g = a[1], b = a[0];
    ans += b;
    r -= b;
    g -= b;
    ans += r / 3 + g / 3;
    if (r % 3 == 2 && g % 3 == 2 && b > 0) ans++;
    cout << ans;
  }
}

```