#include <bits/stdc++.h>
using namespace std;
long long poww(long long x, long long y, long long m) {
  x = x % m;
  if (y == 0) {
    return 1;
  }
  if (y % 2 == 0) {
    return poww(x * x, y / 2, m);
  } else {
    return (x * poww(x * x, y / 2, m)) % m;
  }
}
void io();
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int a, b, c;
int first(int x) { return x + (a - x) / 3 + (b - x) / 3 + (c - x) / 3; }
int main() {
  io();
  int i, j;
  cin >> a >> b >> c;
  int mn = min(a, min(b, c));
  int a1 = a / 3 + b / 3 + c / 3;
  int mx = a1;
  for ((i) = 0; (i) < (min(mn, 3) + 1); i = i + 1) {
    mx = max(mx, first(i));
  }
  cout << mx;
  return 0;
}
void io() {}

### 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;
long long poww(long long x, long long y, long long m) {
  x = x % m;
  if (y == 0) {
    return 1;
  }
  if (y % 2 == 0) {
    return poww(x * x, y / 2, m);
  } else {
    return (x * poww(x * x, y / 2, m)) % m;
  }
}
void io();
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int a, b, c;
int first(int x) { return x + (a - x) / 3 + (b - x) / 3 + (c - x) / 3; }
int main() {
  io();
  int i, j;
  cin >> a >> b >> c;
  int mn = min(a, min(b, c));
  int a1 = a / 3 + b / 3 + c / 3;
  int mx = a1;
  for ((i) = 0; (i) < (min(mn, 3) + 1); i = i + 1) {
    mx = max(mx, first(i));
  }
  cout << mx;
  return 0;
}
void io() {}

```