#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1000 * 1001;
const long long INF = 1e18;
const long long mod = 998244353;
long long n, m, k;
long long mark[maxn], a[maxn];
vector<long long> v;
int main() {
  ios::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  long long ans = 0;
  long long g = min(n, min(m, k));
  long long t = 0;
  for (long long h = max(g - 7, t); h <= g; h++) {
    long long anss = (k - h) / 3 + (m - h) / 3 + (n - h) / 3 + (h);
    ans = max(ans, anss);
  }
  cout << ans;
}

### Prompt

Your challenge is to write 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;
const long long maxn = 1000 * 1001;
const long long INF = 1e18;
const long long mod = 998244353;
long long n, m, k;
long long mark[maxn], a[maxn];
vector<long long> v;
int main() {
  ios::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  long long ans = 0;
  long long g = min(n, min(m, k));
  long long t = 0;
  for (long long h = max(g - 7, t); h <= g; h++) {
    long long anss = (k - h) / 3 + (m - h) / 3 + (n - h) / 3 + (h);
    ans = max(ans, anss);
  }
  cout << ans;
}

```