Problem:
Score : 200 points

Problem Statement

We have a board with H horizontal rows and W vertical columns of squares.
There is a bishop at the top-left square on this board.
How many squares can this bishop reach by zero or more movements?

Here the bishop can only move diagonally.
More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds:

r_1 + c_1 = r_2 + c_2

r_1 - c_1 = r_2 - c_2

For example, in the following figure, the bishop can move to any of the red squares in one move:

Constraints

1 \leq H, W \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H \ W

Output

Print the number of squares the bishop can reach.

Sample Input 1

4 5

Sample Output 1

10

The bishop can reach the cyan squares in the following figure:

Sample Input 2

7 3

Sample Output 2

11

The bishop can reach the cyan squares in the following figure:

Sample Input 3

1000000000 1000000000

Sample Output 3

500000000000000000

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

#include <bits/stdc++.h>
using namespace std;
#define ll long long 
#define endl "\n"
#define pb push_back
#define f(i,n) for(i=0;i<n;i++)
#define F(i,a,b) for(i=a;a<=b;i++)
#define arr(a,n) for( i=0;i<n;i++)cin>>a[i];
#define fi first
#define se second
#define mp make_pair
#define mod 1000000007
#define YES cout<<"YES"<<endl;
#define Yes cout<<"Yes"<<endl;
#define NO cout<<"NO"<<endl;
#define No cout<<"No"<<endl;
#define yes cout<<"yes"<<endl;
#define no cout<<"no"<<endl;
#define vi vector<ll>
#define ed end()
#define bg begin()
#define sz size()
#define ln length()
#define s() sort(a,a+n);
#define sr() sort(a,a+n,greater<ll>());
#define v()  sort(v.begin(),v.end());
#define vr() sort(v.begin(),v.end(),greater<ll>());
#define mod 1000000007
#define fast() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
ll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);}
ll power(ll x,ll y,ll p){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}
int main() {
   /* #ifndef ONLINE_JUDGE
        // for getting input from input.txt
        freopen("input.txt", "r", stdin);
        // for writing output to output.txt
        freopen("output.txt", "w", stdout);
    #endif*/
        /* 
        Ofcourse it's Hard.
        It's supposed to be Hard.
        If it's easy everyone would do it.
        HARD IS WHAT MAKES IT GREAT

        YESTERDAY U SAID TOMORROW

        SLOWLY BECOMING THE PERSON I 
        SHOULD HAVE BEEN A LONG TIME AGO

        SAME TASK CAN'T BE FOUND DIFFICULT TWICE

        BTBHWSITW

        SPRH TU KAB P
        
        IT ALWAYS SEEMS IMPOSSIBLE
        UNTIL ITS DONE ->WILLIAM LIN


        */

fast();
//ll t;cin>>t>>while(t--)
{
	 unsigned ll h,w;
	cin>>h>>w;
	unsigned ll x=ceil(double(h)/double(2.0));
	unsigned ll y=ceil(double(w)/double(2.0));
	if(w%2!=0)
	 {
	 	cout<<(x*y)+(h-x)*(y-1)<<endl;}
	else
	{
		cout<<(x*y)+(h-x)*(y)<<endl;
	}


    
}
return 0;
}