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: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <numeric>
//#include <stdio.h>
//#include <stack>
//#include <queue>
//#include <cstdio>
#include <cmath>
#include <iterator>
#include <map>
#include <fstream>
//#include <list>
#include <iomanip>
//#include <stdlib.h>
//#include <cctype>

using namespace std;

const int MAX_N = 1000000;
#define IOS ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define prec std::cout << std::fixed << std::setprecision(15);
#define endl "\n"
#define MOD 1000000007
#define Int int64_t
#define PI 3.14159265358979
#define rep(i, n) for (int i = 0; i < int(n); i++)
#define ssort(z) sort(z.begin(), z.end())
#define rsort(z) sort(z.rbegin(), z.rend())
#define eerase(z) z.erase(unique(z.begin(), z.end()), z.end())
#define ccnt(z, w) count(z.begin(), z.end(), w)
#define rep(i,a,n) for(Int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(Int (i)=(a); (i)<=(n); (i)++)

template <typename T>
void printV(const std::vector<T>& v, const char * const separator = " ")
{
    if(!v.empty())
    {
        std::copy(v.begin(),
                  --v.end(),
                  std::ostream_iterator<T>(std::cout, separator));
        std::cout << v.back() << "\n";
    }
}


Int gcd(Int a, Int b) {
    return b != 0 ? gcd(b, a % b) : a;
}

Int lcm(Int a, Int b) {
    return a / gcd(a, b) * b;
    //a*bは64bit integer overflow
}

int Max(int a, int b, int c) {
    int temp = max(a, b);
    return max(temp, c);
}

int Min(int a, int b, int c) {
    int temp = min(a, b);
    return min(temp, c);
}

bool integer(double num) {
    return floor(num) == num;
}

Int fact(int num) {
    if (num == 0)
        return 1;
    else
        return num * fact(num - 1);
}

void y() {
    cout << "Yes" << endl;
    exit(0);
}

void n() {
    cout << "No" << endl;
    exit(0);
}

void Y() {
    cout << "YES" << endl;
    exit(0);
}

void N() {
    cout << "NO" << endl;
    exit(0);
}

Int yakusu(int n) {
    int cnt = 0;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            cnt++;
            if (i * i != n) cnt++;
        }
    }
    return cnt;
}

/*   kokomade  */

int cnt = 0;
bool is_prime[ 50000 + 1 ];
vector<int> P;
void Eratosthenes( const int N )
{
    for( int i = 0; i <= N; i++ )
    {
        is_prime[ i ] = true;//初期化
    }
    for( int i = 2; i <= N; i++ )
    {
        if( is_prime[ i ] )
        {
            for( int j = 2 * i; j <= N; j += i )
            {
                is_prime[ j ] = false;
            }
            P.emplace_back( i );
        }
    }
}
bool compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a.second != b.second){
        return a.second < b.second;
    }else{
        return a.first < b.first;
    }
}
vector< int64_t > divisor(int64_t n) {
    vector< int64_t > ret;
    for(int64_t 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);
}


void euclid(Int N, Int a, Int b)
{
    for (Int d = a+b; d <= N; d += b) {
        //cout << b << ", " << d << endl;
        euclid(N, b, d);
    }
}

Int fact_mod(Int n, Int mod) {
    Int f = 1; repq(i,2,n) f = f * (i % MOD) % MOD;
    return f;
}

// 繰り返し二乗法 (modの世界での累乗)
Int mod_pow(Int x, Int n, Int mod) {
    Int res = 1;
    while(n > 0) {
        if(n & 1) res = (res * x) % mod;
        x = (x * x) % mod;
        n >>= 1;
    }
    return res;
}

// 組み合わせ nCr を求める
Int combination_mod(Int n, Int r, Int mod) {
    if(r > n-r) r = n-r;
    if(r == 0) return 1;
    Int a = 1;
    rep(i, 0, r) a = a * ((n-i) % mod) % mod;
    Int b = mod_pow(fact_mod(r, mod), mod-2, mod);
    return (a % mod) * (b % mod) % mod;
}
int main() {
    //COMinit();
    IOS;
    prec;
    Int a,b,n,k,x,y=0,ans=0;
    string str,s,t,u;
    cin>>a>>b;
    x=(a+1)/2;
    if(a%2==0){
        ans=x*b;
    }else{
        if(b%2==0){
            ans=(x*b/2)+(x-1)*b/2;
        }else{
            ans=x*(b+1)/2+(x-1)*(b-1)/2;
        }
    }
    if (b==1 or a==1)ans=1;
    cout<<ans;


    cout<<endl;
    return 0;
}

/*
テレビも無ェ ラジオも無ェ
自動車もそれほど走って無ェ
ピアノも無ェ バーも無ェ
巡査 毎日ぐーるぐる
朝起ぎで 牛連れで
二時間ちょっとの散歩道
電話も無ェ 瓦斯も無ェ
バスは一日一度来る
俺らこんな村いやだ 俺らこんな村いやだ
東京へ出るだ 東京へ出だなら
銭コァ貯めで 東京でベコ(牛)飼うだ
*/