Problem:
Score : 100 points

Problem Statement

There is a right triangle ABC with ∠ABC=90°.

Given the lengths of the three sides, |AB|,|BC| and |CA|, find the area of the right triangle ABC.

It is guaranteed that the area of the triangle ABC is an integer.

Constraints

1 \leq |AB|,|BC|,|CA| \leq 100

All values in input are integers.

The area of the triangle ABC is an integer.

Input

Input is given from Standard Input in the following format:

|AB| |BC| |CA|

Output

Print the area of the triangle ABC.

Sample Input 1

3 4 5

Sample Output 1

6

This triangle has an area of 6.

Sample Input 2

5 12 13

Sample Output 2

30

This triangle has an area of 30.

Sample Input 3

45 28 53

Sample Output 3

630

This triangle has an area of 630.

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

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using namespace boost::multiprecision;

# define REP(i,n) for (int i=0;i<(n);++i)
# define rep(i,a,b) for(int i=a;i<(b);++i)
# define p(s) std::cout << s ;
# define pl(s)  std::cout << s << endl;
# define all(v) v.begin(),v.end()
# define showVector(v) REP(i,v.size()){p(v[i]);p(" ")} pl("")
template<class T> inline bool chmin(T &a, T b){ if(a > b) { a = b; return true;} return false;}
template<class T> inline bool chmax(T &a, T b){ if(a < b) { a = b; return true;} return false;}
typedef long long int ll;
typedef pair<ll,ll> P_ii;
typedef pair<double,double> P_dd;

template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}

template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
  return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}

template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}

template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){
  for(auto &e:t) fill_v(e,v);
}

ll gcd(ll a, ll b) {
    if(a < b) swap(a,b);
    
    if(b == 0) return a;
    return gcd(b, a % b);
}

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

// 素数判定 O(√n)
bool is_prime(int n){
    for(int i = 2; i * i <= n; i++){
        if(n % i == 0) return false;
    }
    return true;
}

// 約数列挙 O(√n)
vector<ll> divisor(ll n){
    vector<ll> res;
    for(ll i = 1; i * i <= n; i++){
        if(n % i == 0){
            res.push_back(i);
            if(i != n / i) res.push_back(n / i);
        }
    }
    return res;
}

vector<pair<ll, ll> > prime_factorize(ll n) {
    vector<pair<ll, ll> > res;
    for (ll p = 2; p * p <= n; ++p) {
        if (n % p != 0) continue;
        ll num = 0;
        while (n % p == 0) { ++num; n /= p; }
        res.push_back(make_pair(p, num));
    }
    if (n != 1) res.push_back(make_pair(n, 1));
    return res;
}

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x(x%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;
  }
};

const int MOD = 1000000007;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;

void addM(ll &a, ll b) {
    a += b;
    if (a >= MOD) a -= MOD;
}

void mulM(ll &a, ll b) {
    a = ((a%MOD)*(b%MOD))%MOD ;
}

ll powM(ll a,ll b) {
    ll ret = 1;
    ll tmp = a;
    while(b>0) {
        if((b&1)==1) ret = (ret * tmp) % MOD;
        tmp = (tmp * tmp) % MOD;
        b = b >> 1;
    }
    return ret;
}

// mod. m での a の逆元 a^{-1} を計算する
ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);  

  int a, b, c;
  cin >> a >> b >> c;

  int ans = a * b / 2;
  cout << ans << endl;

  return 0;
}