#include <bits/stdc++.h>
using namespace std;
double eps = 1e-9;
int inf = 1000000005;
int mod = 1000000007;
long long INFF = 1000000000000000005ll;
double PI = acos(-1);
void cp() {}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long i, j, k;
  long long t;
  long long r, b, g;
  cin >> r >> g >> b;
  long long ans[3] = {0};
  long long mn;
  mn = min(r, (min(g, b)));
  j = 2;
  mn = min(j, mn);
  long long mx = 0;
  for (i = 0; i <= mn; ++i) {
    ans[i] += i + (r - i) / 3 + (g - i) / 3 + (b - i) / 3;
    mx = max(ans[i], mx);
  }
  cout << mx << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
double eps = 1e-9;
int inf = 1000000005;
int mod = 1000000007;
long long INFF = 1000000000000000005ll;
double PI = acos(-1);
void cp() {}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long i, j, k;
  long long t;
  long long r, b, g;
  cin >> r >> g >> b;
  long long ans[3] = {0};
  long long mn;
  mn = min(r, (min(g, b)));
  j = 2;
  mn = min(j, mn);
  long long mx = 0;
  for (i = 0; i <= mn; ++i) {
    ans[i] += i + (r - i) / 3 + (g - i) / 3 + (b - i) / 3;
    mx = max(ans[i], mx);
  }
  cout << mx << endl;
  return 0;
}

```