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"bits/stdc++.h"
using namespace std;
typedef unsigned long long ll;


using Graph=vector<vector<int>>;//グラフ型
void dfs(const Graph &G,int v/*根*/,vector<bool> &seen){
    seen[v]=true;  //vを訪問済にする
    for(auto next_v:G[v]){ //訪問済にしたvとつながっている場所を順番に探す
        if(seen[next_v]) continue; //もし探索済みならスルー
        dfs(G,next_v,seen); //再帰的に探索
    }
}

void bfs(const Graph &G,queue<int> &que/*発見したが訪問はしていない*/,vector<int> &dist/*開始地点からの距離*/,int s/*スタート地点*/){
    
    while(!que.empty()){
        int v=que.front();//queのうち先頭のものからスタート
        que.pop();//先頭を削除
        for(int nv :G[v]){//vとつながっている点を列挙
            if(dist[nv]!=-1) continue;//発見済(=距離が格納されている)であればスルー
            dist[nv]=dist[v]+1;//vとつながっている点に距離の情報を追加
            que.push(nv);//vとつながっている点を発見済にする
        }
    }
}

unsigned int to_int(char c){
    return c-48;
}

int main(void){
    ll h,w;
    cin>>h>>w;
    if(h==1||w==1){
        cout<<1;
    }
    else if((h*w)%2==1){
        cout<<(h*w+1)/2;
    }
    else{
        cout<<h*w/2;
    }
}