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 Ruby (2.3.3) (judge verdict: Accepted)

h,w,k = gets.split.map &:to_i
s = Array.new(h){ gets.chomp.chars.map &:to_i }

cs = Array.new(h){ [0] * (w+1) }
h.times do |y|
  s[y].each_with_index do |v, i|
    cs[y][i + 1] = cs[y][i] + s[y][i]
  end
end

# p [h,w,k,s,cs]
# p cs.map{|a|a[-1]}

# 縦の分割パターンがbitのとき
# left,rightで分割したときの
# ホワイトチョコの最大値を返す
def count(bit,left,right,cs,h)
  ans = 0
  cnt = 0
  h.times do |y|
    # p [left,right,cs]
    # exit
    cnt += cs[y][right] - cs[y][left]
    if bit[y] == 1 # 分割あり
      ans = cnt if ans < cnt
      cnt = 0
    end
  end
  ans = cnt if ans < cnt
  return ans
end

# p count(0, 0, 3, cs, h)
$c = 0
ans = 10**9
(2**(h-1)).times do |bit|
  # p [bit]
  left = 0
  right = 1
  div = bit.to_s(2).count("1") # 分割数
  while right <= w
    ret = count(bit, left, right, cs, h)
    if ret > k # && left < right # 超える前のrightを採用
      if left + 1 == right
        div = 10**9
        break # whileを抜けて、left=0, right=1からやり直し
      else
        div += 1
        left = right - 1
      end
    else
      right += 1
    end
    # p [bit,left,right,ret,k,div]
  end
  ans = div if ans > div
end

puts ans