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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

# -*- coding: utf-8 -*-

# H <= 10であるため2^(H-1)を全探索し、
# W <= 1000 については貪欲に左から探索していけば良い

'''
4 10 4
1110010010
1000101110
0011101001
1101000111
'''

import numpy as np


def main():
    H, W, K = map(int, input().split())

    data = np.array([[None] * W for _ in range(H)])

    for i in range(H):
        temp = input()
        for j in range(W):
            data[i, j] = int(temp[j])

    current_best = np.inf

    # Hに関する全ての場合を試す
    for cnt in range(2 ** (H-1)):
        # bitで1の場所が切れ目とする
        option_h = np.array([int(s) for s in list(bin(cnt))[2:][::-1]])
        kireme = np.where(option_h == 1)[0]

        # 各グループについて1の数を集計する
        li = 0
        g = np.array([[None] * W for _ in range(len(kireme) + 1)])
        if len(kireme) == 0:
            g[0] = data[li:, :].sum(axis=0)
        else:
            for idx, j in enumerate(kireme):
                g[idx] = data[li:j+1, :].sum(axis=0)
                li = j + 1
            g[-1] = data[li:, :].sum(axis=0)

        # 左から順に足していく
        score = len(kireme)
        if score > current_best:
            # すでにこの時点で回数かかっていれば探索打ち切り
            continue

        res = np.zeros(len(kireme) + 1)
        for col in range(W):
            res = res + g[:, col]
            # どこか1つのグループでもKを超えてしまったらその左で切る
            if (res > K).any():
                score += 1
                # 最小を求めたいので、それまでの結果と同じ値になってしまったら終了
                if score >= current_best:
                    break
                res = g[:, col]
                # 現在の列だけで条件を満たせないのであれば不可能なので終了
                if (res > K).any():
                    score = np.inf
                    break
        # スコアの更新
        # print(cnt, score)
        if score < current_best:
            current_best = score

    print(current_best)


if __name__ == '__main__':
    main()