Problem:
Score: 100 points

Problem Statement

There is a train going from Station A to Station B that costs X yen (the currency of Japan).

Also, there is a bus going from Station B to Station C that costs Y yen.

Joisino got a special ticket. With this ticket, she can take the bus for half the fare if she travels from Station A to Station B by train and then travels from Station B to Station C by bus.

How much does it cost to travel from Station A to Station C if she uses this ticket?

Constraints

1 \leq X,Y \leq 100

Y is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

If it costs x yen to travel from Station A to Station C, print x.

Sample Input 1

81 58

Sample Output 1

110

The train fare is 81 yen.

The train fare is 58 ⁄ 2=29 yen with the 50% discount.

Thus, it costs 110 yen to travel from Station A to Station C.

Sample Input 2

4 54

Sample Output 2

31

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

#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define REP(i, n) for(ll i = 1; i < (ll)(n); i++)
#define P pair<ll, ll>
typedef long long ll;
const ll mod = 1e9+7;

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;}

//繰り返し自乗法による累乗計算
//a**n
ll pow(ll a, ll n){
  if(n == 0) return 1;
  if(n%2 == 0){
    ll res = pow(a, n/2);
    return res * res;
  }
  return a * pow(a, n-1);
}
//繰り返し自乗法によるmod累乗計算
//a**n (mod)
ll modpow(ll a, ll n){
  if(n == 0) return 1;
  if(n%2 == 0){
    ll res = modpow(a, n/2);
    return res * res % mod;
  }
  return a * modpow(a, n-1) % mod;
}
//組み合わせ
//nCa
ll com(ll n, ll a){
  ll x = 1, y = 1;
  rep(i,a){
    x *= n-i;
    y *= i+1;
  }
  ll res = x / y;
  return res;
}
//組み合わせ(mod)
//nCa (mod)
ll modcom(ll n, ll a){
  ll x = 1, y = 1;
  rep(i, a){
    x = x * (n-i) % mod;
    y = y * (i+1) % mod;
  }
  ll ans = x * modpow(y, mod-2) % mod;
  return ans;
}
//gcd 最大公約数
ll gcd(int x, int y){
  if(x % y == 1) return y;
  return gcd(y, x%y);
}
//lcm 最小公倍数
ll lcm(int x, int y){
  return x * y / gcd(x, y);
}
// n個の最大公約数
ll ngcd(vector<int> a){
  int res;
  res = a[0];
  for(int i=1; i<a.size() && res!=1; i++){
    res = gcd(res, a[i]);
  }
  return res;
}
// n個の最小公倍数
ll nlcm(vector<int> a){
  int res;
  res = a[0];
  REP(i, a.size()){
    res = lcm(res, a[i]);
  }
  return res;
}
// 素数判定
bool isprime(int x){
  int i;
  if(x < 2) return false;
  else if(x == 2) return true;
  if(x % 2 == 0) return false;
  for(i = 3; i*i <= x; i += 2){
    if(x % i == 0) return false;
  }
  return true;
}
// 桁和
int digsum(int x){
  int res = 0;
  while(x > 0){
    res += x % 10;
    x /= 10;
  }
  return res;
}


int main(){
  int x, y;
  cin >> x >> y;
  int ans = x + y/2;
  cout << ans << endl;
}