Problem:
Score : 300 points

Problem Statement

We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is ., and black if c_{i,j} is #.

Consider doing the following operation:

Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.

You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

Constraints

1 \leq H, W \leq 6

1 \leq K \leq HW

c_{i,j} is . or #.

Input

Input is given from Standard Input in the following format:

H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W}

Output

Print an integer representing the number of choices of rows and columns satisfying the condition.

Sample Input 1

2 3 2
..#
###

Sample Output 1

5

Five choices below satisfy the condition.

The 1-st row and 1-st column

The 1-st row and 2-nd column

The 1-st row and 3-rd column

The 1-st and 2-nd column

The 3-rd column

Sample Input 2

2 3 4
..#
###

Sample Output 2

1

One choice, which is choosing nothing, satisfies the condition.

Sample Input 3

2 2 3
##
##

Sample Output 3

0

Sample Input 4

6 6 8
..##..
.#..#.
#....#
######
#....#
#....#

Sample Output 4

208

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	static int[][] map;
	static int[][] directions8= {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
	static int[][] directions4= {{-1,0},{1,0},{0,-1},{0,1}};
	static int ans;
	static int H;
	static int W;
	static int K;

	static boolean[][] c;

	public static void main(String[] args) throws IOException {
		//		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//		String W	 = in.readLine();
		//		int num = Integer.parseInt(W.split(" ")[0]);

		Scanner sc = new Scanner(System.in);
		H = sc.nextInt();
		W = sc.nextInt();
		K = sc.nextInt();

		c  = new boolean[H][W];
		int blackNum = 0;

		for(int i = 0; i<H; i++) {
			String str = sc.next();
			for(int j = 0;j<W ; j++) {
				if(str.charAt(j)=='#') {
					blackNum++;

					c[i][j]=true;
				}
			}
		}

		ans =0;

		solve(new boolean[H+W], 0);


		System.out.println(ans);
	}



	static void solve(boolean[] lines, int idx) {
		if(idx == lines.length) {
			evaluate(lines);
			return;
		}else {
			lines[idx] = false;
			solve(lines, idx+1);
			lines[idx] = true;
			solve(lines, idx+1);
		}
		return;
	}

	static void evaluate(boolean[] lines) {
		boolean[][] tmp = new boolean[H][W];
		for(int i = 0;i<H ; i++) {
			for(int j = 0;j<W ; j++) {
				if(c[i][j]) {
					tmp[i][j]=true;
				}
			}
		}
		for(int i = 0;i<H ; i++) {
			if(lines[i]) {
				Arrays.fill(tmp[i], false);
			}
		}
		for(int i = H;i<W+H ; i++) {
			if(lines[i]) {
				for(int j = 0;j<H ; j++) {
					tmp[j][i-H] = false;
				}
			}
		}
		int tmpBlack = 0;
		for(int i = 0;i<H ; i++) {
			for(int j = 0;j<W ; j++) {
				if(tmp[i][j]) {
					tmpBlack++;
				}
			}
		}
		if(tmpBlack == K) {
			ans++;
		}
		return;
	}

	//BFS用に二つの配列を足し算する
	static int[] addArrayElms(int[] a, int[] b) {
		int[] c = new int[a.length];
		for(int i = 0; i < a.length; i++) {
			c[i] = a[i] + b[i];
		}
		return c;
	}

	//	//二分探索
	//k <= num となる最小の配列要素kのインデックスを返す
	static private int binarySearch(long num, long[] orderedArray){
		int lowerBorder = -1;
		int upperBorder = orderedArray.length;
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedArray[mid]<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//二分探索
	//k <= num となる最小のList要素kのインデックスを返す
	static private int binarySearch(long num, ArrayList<Long> orderedList){
		int lowerBorder = -1;
		int upperBorder = orderedList.size();
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedList.get(mid)<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//aとbの最小公倍数を求める
	public static int gcd(int a, int b) {
		return b == 0 ? a: gcd(b, a % b);
	}
	public static long gcd(long a, long b) {
		return b == 0 ? a: gcd(b, a % b);
	}
}