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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import qualified Data.Vector.Unboxed as U
import qualified Data.ByteString.Char8 as C
import Debug.Trace
import Data.List
import Control.Monad
import Data.Ord
import Data.Maybe
import Numeric
import Data.Char
import Text.Printf

r :: Int -> IO (U.Vector Int)
r n = U.unfoldrN n (C.readInt . C.dropWhile (==' ')) <$> C.getLine


r' :: IO [Int]
r'  =   unfoldr    (C.readInt . C.dropWhile (==' ')) <$> C.getLine


main :: IO ()
main = do
    [h,w,k] <- r'
    choco <- replicateM h (map (read.pure) <$> getLine) :: IO[[Int]]
    let sepH = seps (h-1)
    let vert = map (sepWLO k) $ filter (prep k) $ map (map (foldzip sum) . (\s->sepBy s choco)) sepH
    print $ minimum (zipWith (+) (map sum sepH) vert)


sepWLO lim xxs = sepWL lim (replicate (length xxs) 0) 0 xxs

prep :: Int -> [[Int]] -> Bool
prep lim xxs = maximum (map maximum xxs) <= lim

sepWL lim whites seps ([]:xxs) = seps
sepWL lim whites seps xxs = let zipped = zipWith (+) (map head xxs) whites in
    if maximum zipped > lim
    then sepWL lim (map head xxs) (seps+1) (map tail xxs)
    else sepWL lim  zipped         seps    (map tail xxs)



foldzip :: ([Int] -> Int) -> [[Int]] -> [Int]
foldzip f xs
    | null (head xs) = []
    | otherwise = f (map head xs):foldzip f (map tail xs)

sepBy :: [Int] -> [a] -> [[a]]
sepBy [] xs = [xs]
sepBy (0:sp) (x:xs) = (x:) `onHeadOf` (sepBy sp xs)
sepBy (1:sp) (x:xs) = [x]:(sepBy sp xs)

f `onHeadOf` (x:xs) = (f x):xs

seps :: Int -> [[Int]]
seps n = map ((map (read.pure)).(printf ("%0"++show n++"b"))) [0..(2^n :: Int)-1]