Problem:
Score : 100 points

Problem Statement

We have A apples and P pieces of apple.

We can cut an apple into three pieces of apple, and make one apple pie by simmering two pieces of apple in a pan.

Find the maximum number of apple pies we can make with what we have now.

Constraints

All values in input are integers.

0 \leq A, P \leq 100

Input

Input is given from Standard Input in the following format:

A P

Output

Print the maximum number of apple pies we can make with what we have.

Sample Input 1

1 3

Sample Output 1

3

We can first make one apple pie by simmering two of the three pieces of apple. Then, we can make two more by simmering the remaining piece and three more pieces obtained by cutting the whole apple.

Sample Input 2

0 1

Sample Output 2

0

We cannot make an apple pie in this case, unfortunately.

Sample Input 3

32 21

Sample Output 3

58

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define FOR(i, m, n) for (int i = m; i < (n); i++)
#define RFOR(i, m, n) for (int i = (m - 1); i >= 0; i--)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, n, 0)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const ll INF = 1000000000000000000LL;

struct mint {
  ll x;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return (*this) *= a.inv(); }
  mint operator/(const mint a) const {
    mint res(*this);
    return res /= a;
  }
};

mint factorial(ll n) {
  mint answer = 1;
  while (n > 1) {
    answer *= n;
    n--;
  }
  return answer;
}

mint combination(ll n, ll r) {
  return factorial(n) / (factorial(r) * factorial(n - r));
}

mint permutation(ll n, ll r) { return factorial(n) / factorial(n - r); }

// setprecision(8) 桁数指定
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }

ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

template <typename T>
void remove(std::vector<T>& vector, unsigned int index) {
  vector.erase(vector.begin() + index);
}

map<ll, ll> prime_factor(ll n) {
  map<ll, ll> ret;
  for (ll i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if (n != 1) ret[n] = 1;
  return ret;
}
// for (auto p : prime_factor(n)) {
//	while (p.second--) {
//		cout << p.first << endl;
//	}
//}

vector<ll> divisor(ll n) {
  vector<ll> ret;
  for (ll i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}

bool compare_by_b(pair<ll, ll> a, pair<ll, ll> b) {
  if (a.second != b.second) {
    return a.second < b.second;
  } else {
    return a.first < b.first;
  }
}

vector<ll> Eratosthenes(const ll N) {
  bool is_prime[1000000 + 1];
  vector<ll> P;
  for (ll i = 0; i <= N; i++) {
    is_prime[i] = true;
  }
  for (ll i = 2; i <= N; i++) {
    if (is_prime[i]) {
      for (ll j = 2 * i; j <= N; j += i) {
        is_prime[j] = false;
      }
      P.emplace_back(i);
    }
  }
  return P;
}

typedef vector<int> ivec;
typedef vector<string> svec;
typedef vector<ll> lvec;

int main() {
  int a, p;
  cin >> a >> p;
  int ans = (a * 3) / 2 + (p / 2);
  cout << ans << endl;
  return 0;
}