#include <bits/stdc++.h>
using namespace std;
int maxx1(int a, int b, int c) {
  int ret = 1;
  a -= 1;
  b -= 1;
  c -= 1;
  ret += a / 3;
  ret += b / 3;
  ret += c / 3;
  return ret + min(a % 3, min(b % 3, c % 3));
}
int maxx2(int r, int g, int b) {
  int ret = 2;
  r -= 2;
  g -= 2;
  b -= 2;
  ret += r / 3;
  ret += g / 3;
  ret += b / 3;
  return ret;
}
int do1(int a, int b, int c) {
  int ret = 1;
  a -= 1;
  b -= 1;
  c -= 1;
  ret += max(maxx1(a, b, c), maxx2(a, b, c));
  return ret;
}
int do2(int r, int g, int b) {
  int ret = 2;
  r -= 2;
  g -= 2;
  b -= 2;
  ret += r / 3;
  ret += g / 3;
  ret += b / 3;
  ret += min(r % 3, min(g % 3, b % 3));
  return ret;
}
int do3(int r, int g, int b) {
  int ret = 3;
  r -= 3;
  g -= 3;
  b -= 3;
  ret += r / 3;
  ret += g / 3;
  ret += b / 3;
  return ret;
}
int main() {
  int r, g, b;
  while (~scanf("%d%d%d", &r, &g, &b)) {
    if (min(r, min(g, b)) % 3 == 0 && min(r, min(g, b)) >= 3) {
      int x = min(r, min(g, b)) / 3 - 1;
      int ans = 3 * x;
      r -= 3 * x;
      g -= 3 * x;
      b -= 3 * x;
      ans += max(do1(r, g, b), max(do2(r, g, b), do3(r, g, b)));
      printf("%d\n", ans);
      continue;
    }
    int ans = 0;
    int mi = min(r, min(g, b));
    int x = mi / 3;
    ans += 3 * x;
    r -= 3 * x;
    g -= 3 * x;
    b -= 3 * x;
    if (min(r, min(g, b)) == 1) {
      ans += r / 3;
      ans += g / 3;
      ans += b / 3;
      ans += min(r % 3, min(g % 3, b % 3));
      printf("%d\n", ans);
      continue;
    }
    if (min(r, min(g, b)) == 2) {
      ans += max(maxx2(r, g, b), maxx1(r, g, b));
      printf("%d\n", ans);
      continue;
    }
    if (min(r, min(g, b)) == 0) {
      ans += r / 3;
      ans += g / 3;
      ans += b / 3;
      printf("%d\n", ans);
      continue;
    }
  }
  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;
int maxx1(int a, int b, int c) {
  int ret = 1;
  a -= 1;
  b -= 1;
  c -= 1;
  ret += a / 3;
  ret += b / 3;
  ret += c / 3;
  return ret + min(a % 3, min(b % 3, c % 3));
}
int maxx2(int r, int g, int b) {
  int ret = 2;
  r -= 2;
  g -= 2;
  b -= 2;
  ret += r / 3;
  ret += g / 3;
  ret += b / 3;
  return ret;
}
int do1(int a, int b, int c) {
  int ret = 1;
  a -= 1;
  b -= 1;
  c -= 1;
  ret += max(maxx1(a, b, c), maxx2(a, b, c));
  return ret;
}
int do2(int r, int g, int b) {
  int ret = 2;
  r -= 2;
  g -= 2;
  b -= 2;
  ret += r / 3;
  ret += g / 3;
  ret += b / 3;
  ret += min(r % 3, min(g % 3, b % 3));
  return ret;
}
int do3(int r, int g, int b) {
  int ret = 3;
  r -= 3;
  g -= 3;
  b -= 3;
  ret += r / 3;
  ret += g / 3;
  ret += b / 3;
  return ret;
}
int main() {
  int r, g, b;
  while (~scanf("%d%d%d", &r, &g, &b)) {
    if (min(r, min(g, b)) % 3 == 0 && min(r, min(g, b)) >= 3) {
      int x = min(r, min(g, b)) / 3 - 1;
      int ans = 3 * x;
      r -= 3 * x;
      g -= 3 * x;
      b -= 3 * x;
      ans += max(do1(r, g, b), max(do2(r, g, b), do3(r, g, b)));
      printf("%d\n", ans);
      continue;
    }
    int ans = 0;
    int mi = min(r, min(g, b));
    int x = mi / 3;
    ans += 3 * x;
    r -= 3 * x;
    g -= 3 * x;
    b -= 3 * x;
    if (min(r, min(g, b)) == 1) {
      ans += r / 3;
      ans += g / 3;
      ans += b / 3;
      ans += min(r % 3, min(g % 3, b % 3));
      printf("%d\n", ans);
      continue;
    }
    if (min(r, min(g, b)) == 2) {
      ans += max(maxx2(r, g, b), maxx1(r, g, b));
      printf("%d\n", ans);
      continue;
    }
    if (min(r, min(g, b)) == 0) {
      ans += r / 3;
      ans += g / 3;
      ans += b / 3;
      printf("%d\n", ans);
      continue;
    }
  }
  return 0;
}

```