Problem:
Score : 100 points

Problem Statement

We have N+M balls, each of which has an integer written on it.

It is known that:

The numbers written on N of the balls are even.

The numbers written on M of the balls are odd.

Find the number of ways to choose two of the N+M balls (disregarding order) so that the sum of the numbers written on them is even.

It can be shown that this count does not depend on the actual values written on the balls.

Constraints

0 \leq N,M \leq 100

2 \leq N+M

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the answer.

Sample Input 1

2 1

Sample Output 1

1

For example, let us assume that the numbers written on the three balls are 1,2,4.

If we choose the two balls with 1 and 2, the sum is odd;

If we choose the two balls with 1 and 4, the sum is odd;

If we choose the two balls with 2 and 4, the sum is even.

Thus, the answer is 1.

Sample Input 2

4 3

Sample Output 2

9

Sample Input 3

1 1

Sample Output 3

0

Sample Input 4

13 3

Sample Output 4

81

Sample Input 5

0 3

Sample Output 5

3

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;
import java.io.*;

public class Main {
	
	public static String fix(String s, int h) {
		while(s.length() < h - 1)
			s = "0" + s;
		return s;
	}
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int h = sc.nextInt(), w = sc.nextInt(), k = sc.nextInt();
		int[][] mat = new int[h][w];
		for(int i = 0; i < h; i++) {
			String s = sc.next();
			for(int j = 0; j < w; j++) {
				mat[i][j] = Integer.parseInt("" + s.charAt(j));
			}
		}
		
		int min = h - 1 + w - 1;
		big: for(int prm = 0; prm <= (1 << (h - 1)) - 1; prm++) {
			int b = Integer.bitCount(prm);
			String s = fix(Integer.toBinaryString(prm), h);
			int[] count = new int[b + 1];
			
			int c = b;
			int pre = -1;
			for(int i = 0; i < w; i++) {
				int[] newCount = new int[b + 1];
				int idx = 0;
				for(int j = 0; j < h; j++) {
					if(j > 0 && (s.charAt(s.length() - j) == '1'))
						idx++;
					newCount[idx] += mat[j][i];
				}
				boolean flag = true;
				for(int u = 0; u <= b; u++) {
					count[u] += newCount[u];
					if(count[u] > k) {
						flag = false;
						break;
					}
				}
				
				if(!flag) {
					c++;
					if(pre == i - 1) {
						continue big;
					} else {
						pre = i - 1;
					}
					for(int u = 0; u <= b; u++) {
						count[u] = newCount[u];
					}
					flag = true;
					for(int u = 0; u <= b; u++) {
						if(count[u] > k) {
							flag = false;
							break;
						}
					}
					if(!flag) {
						continue big;
					}
				}
			}
			
			min = Math.min(min, c);
		}
		
		pw.println(min);
		pw.flush();
	}

	public static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}