Problem:
Score : 500 points

Problem Statement

We have a chocolate bar partitioned into H horizontal rows and W vertical columns of squares.

The square (i, j) at the i-th row from the top and the j-th column from the left is dark if S_{i,j} is 0, and white if S_{i,j} is 1.

We will cut the bar some number of times to divide it into some number of blocks. In each cut, we cut the whole bar by a line running along some boundaries of squares from end to end of the bar.

How many times do we need to cut the bar so that every block after the cuts has K or less white squares?

Constraints

1 \leq H \leq 10

1 \leq W \leq 1000

1 \leq K \leq H \times W

S_{i,j} is 0 or 1.

Input

Input is given from Standard Input in the following format:

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

Output

Print the number of minimum times the bar needs to be cut so that every block after the cuts has K or less white squares.

Sample Input 1

3 5 4
11100
10001
00111

Sample Output 1

2

For example, cutting between the 1-st and 2-nd rows and between the 3-rd and 4-th columns - as shown in the figure to the left - works.

Note that we cannot cut the bar in the ways shown in the two figures to the right.

Sample Input 2

3 5 8
11100
10001
00111

Sample Output 2

0

No cut is needed.

Sample Input 3

4 10 4
1110010010
1000101110
0011101001
1101000111

Sample Output 3

3

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int h = sc.nextInt();
        int w = sc.nextInt();
        int k = sc.nextInt();
        
        // input:   3 5 4
        //          11100
        //          10001
        //          00111
        //
        // process: boardに切り込みを入れて
        //          どの部分も'1'がk個以下であるようにする
        //
        // output:  2 (縦1,横3の2箇所に切り込みを入れる)
        
        // board[i][j]: チョコレートの模様
        char[][] board = new char[h][w];
        for (int i = 0; i < h; i++) {
            String str = sc.next();
            for (int j = 0; j < w; j++) {
                board[i][j] = str.charAt(j);
            }
        }
        
        // bit: 横線の引き方のbit全探索
        // min: cutの最小値
        int min = (h-1)*(w-1);
        for (int bit = 0; bit < 1<<(h-1); bit++) {
            // group[i]: i行が属するグループ番号
            // bitに基づいて各行をグループ分けする
            int[] group = new int[h];
            int g = 0;
            group[0] = g;
            for (int i = 0; i < h-1; i++) {
                // bitXが1なら次のグループにする
                if ((bit & (1<<i)) != 0) {
                    g++;
                }
                group[i+1] = g;
            }
            
            // cut: 横線の本数( + 縦線の本数)
            int cut = g;
            
            // count[i]: グループiの現在の'1'の個数
            // last: 最後に引いた縦線の位置
            // ok: 1列だけでk個を超すグループが存在しないならtrue
            int[] count = new int[g+1];
            int last = 0;
            boolean ok = true;
            
            // board -> cut
            // cut += 縦線の本数
            for (int i = 0; i < w; i++) {
                for (int j = 0; j < h; j++) {
                    if (board[j][i] == '1') {
                        count[group[j]]++;
                    }
                    
                    // '1'の個数がkを超えるグループがある場合、
                    // 今の列の前でカットし、その列は数え直し
                    if (count[group[j]] > k) {
                        if (last != i) {
                            cut++;
                            Arrays.fill(count, 0);
                            last = i;
                            i--;
                            break;
                        } else {
                            ok = false;
                        }
                    }
                }
            }
            
            if (ok) {
                min = Math.min(min, cut);
            }
        }
        System.out.println(min);
    }
}