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.
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
#pragma GCC optimize("Ofast")
using namespace std;
void fastio() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long dp[101][1001], ans1 = 1e9, dp1[1000001];
long long num[10];
long long a[1000000];
signed main() {
  fastio();
  long long t = 1;
  while (t--) {
    long long r, g, b, x = 0, y = 0, z = 0, qq = 0;
    cin >> r >> g >> b;
    if ((r == 0 && (g < 3 && b < 3)) || (g == 0 && (r < 3 && b < 3)) ||
        (b == 0 && (g < 3 && r < 3))) {
      cout << 0;
      return 0;
    }
    long long ans = 0;
    if (r % 3 == 0)
      ans += r / 3;
    else {
      ans += (r / 3);
      long long z1 = r / 3;
      x = (r - (3 * z1));
    }
    if (g % 3 == 0)
      ans += g / 3;
    else {
      ans += g / 3;
      long long z2 = g / 3;
      y = (g - (z2 * 3));
    }
    if (b % 3 == 0)
      ans += b / 3;
    else {
      ans += b / 3;
      long long z3 = b / 3;
      z = b - (z3 * 3);
    }
    if (x == 2 && y == 2 && z == 0 && b != 0) qq++;
    if (y == 2 && z == 2 && x == 0 && r != 0) qq++;
    if (x == 2 && z == 2 && y == 0 && g != 0) qq++;
    ans += min({x, y, z});
    if (qq) ans++;
    cout << ans;
  }
}
