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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.PrintStream;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception{
		new Main(new Scanner(System.in), System.out, System.err).exec();
	}
	public final Scanner sc;
	public final PrintStream out, err;
	public Main(final Scanner sc, final PrintStream out, final PrintStream err) {
		this.sc = sc;this.out = out;this.err = err;
	}
	static public final long pair(int a, int b) {return (((long)a) << 32) + b;}
	static public final int getPair1(long v) {return (int)(v >> 32);}
	static public final int getPair2(long v) {return (int)(v & 0xFFFFFFFF);}

	static public final long quad(long a, long b, long c, long d) {
		return ((a&0xFFFF)<< 48) + ((b&0xFFFF)<< 32) + ((c&0xFFFF)<< 16) +(d&0xFFFF);}
	static public final int getQuad1(long v) {return (int)( v >> 48          );}
	static public final int getQuad2(long v) {return (int)((v >> 32) & 0xFFFF);}
	static public final int getQuad3(long v) {return (int)((v >> 16) & 0xFFFF);}
	static public final int getQuad4(long v) {return (int)( v        & 0xFFFF);}

	static public final long record(int e, int m, int i) {
		return quad(-1, e, m, i);
	}
	static public final int re(long record) {
		return getQuad2(record);
	}
	static public final int rm(long record) {
		return getQuad3(record);
	}
	static public final int ri(long record) {
		return getQuad4(record);
	}

	//------------------------------------------

	private int p;
	private int[] inv;
	private int[] fact;
	private int[] finv;
	public final int mod(long v) {
		return mod(v, p);
	}
	static public final int mod(long v, int M) {
		v %= M;
		return (int)(v < 0 ? v+M : v);
	}
	/**
	 * a * b (mod p)
	 * 事前にinitFactを行うこと。
	 */
	public final int mul(long a, long b) {
		return mod(a*b);
	}
	public final void initFact(double limit, double p) {
		initFact((int)limit, (int)p);
	}
	/**
	 * 集合数nの組み合わせ計算を高速化するため、事前に計算を行う。
	 * 組み合わせ数はpで割った余りを算出する。
	 */
	public final void initFact(int n, int p) {
		this.p = p;
		this.inv = new int[n+1];
		this.fact = new int[n+1];
		this.finv = new int[n+1];
		fact[1] = finv[1] = inv[0] = inv[1] = 1;
		for(int i=2;i<=n;i++) {
			int a = p / i;
			int b = p % i;
			inv[i]  = mul(-a, inv[b]);
			fact[i] = mul(fact[i-1], i);
			finv[i] = mul(finv[i-1], inv[i]);
		}
	}
	/**
	 * mCnを計算する。事前にinitFactを実行すること。
	 */
	public final int comb(int m, int n) {
		return mul(mul(fact[m], finv[n]), finv[m-n]);
	}
	/**
	 * a^b (mod M)を計算する。
	 */
	public static long pow(long a, long b, int M) {
		long ret = 1;
		long tmp = a;
		while (b > 0) {
			if ((b & 1) == 1) {
				ret = (ret * tmp) % M;
			}
			tmp = (tmp * tmp) % M;
			b = b >> 1;
		}
		return ret;
	}
	/**
	 * nCk (mod M)を計算する。
	 */
	public static int nCk(int n, int k, int M) {
		long ret = 1;
		int min = Math.min(k, n - k);
		// 分母
		for (int i = 1; i <= min; i++) {
			// a^-1 ≡ a^M-2  ※フェルマーの小定理
			ret = (ret * pow(i, M - 2, M)) % M;
		}
		// 分子
		for (int i = n - min + 1; i <= n; i++) {
			ret = (ret * i) % M;
		}
		return (int) ret;
	}
	//------------------------------------------
	static long lcm (long a, long b) {
		long temp;
		long c = a;
		c *= b;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return (long)(c/b);
	}

	//------------------------------------------

	static class DisjointSetUnion {
		public int[] parent;
		public int[] weight;
		public int count;

		public DisjointSetUnion(int nodes) {
			count = nodes;
			parent = new int[nodes];
			weight = new int[nodes];
			for (int i = 0; i < nodes; i++) {
				parent[i] = i;
				weight[i] = 1;
			}
		}
		//"find"
		public int root(int p) {
			while (p != parent[p]) {
				p = parent[p];
			}
			return p;
		}

		//"union"
		public void connect(int p, int q) {
			int rootP = root(p);
			int rootQ = root(q);
			if (rootP == rootQ) return;
			if (weight[rootP] < weight[rootQ]) {
				parent[rootP] = rootQ;
				weight[rootQ] += weight[rootP];
			} else {
				parent[rootQ] = rootP;
				weight[rootP] += weight[rootQ];
			}
			count--;
		}

		public boolean connected(int p, int q) {
			return root(p) == root(q);
		}
	}
	static class IntList{
		int tail = 0;
		final int[] element;
		public IntList(int cap) {
			element = new int[cap];
		}
		public int size() {
			return tail;
		}
		public int get(int index) {
			return element[index];
		}
		public void reset() {
			tail = 0;
		}
		public void add(int v) {
			element[tail++] = v;
		}
	}

	public void exec() throws Exception{
		final long H = sc.nextLong();
		final long W = sc.nextLong();
		if(W%2 == 0) {
			// セルが偶数の場合は半分
			out.println(H*W/2);
		}else if(H%2 == 0) {
			// セルが偶数の場合は半分
			out.println(H*W/2);
		}else {
			// 両方奇数
			out.println(H*(W-1)/2 + H/2 + 1);
		}


	}
}