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 Fortran (gfortran v4.8.4) (judge verdict: Runtime Error)

program main
    implicit none
    integer(8):: h,w,k
    integer(8):: i,j
    integer(8), allocatable:: c(:,:), now(:)
    integer(8), parameter:: inf = 1001001001
    character(1), allocatable:: s(:,:)
    read*, h,w,k
 
    allocate(s(w,h))

    block
        character(:), allocatable:: input_s
        allocate(character(w)::input_s)

        do i=1,h
            read*, input_s(:)
            do j=1,w
                s(j,i) = input_s(j:j)
            end do
        end do
    end block

    block
        integer(8):: div, g, ints, num
        integer(8):: ans = inf
        integer(8), allocatable:: id(:)
        logical:: emp, ok

        allocate(id(h))
        do div=1, lshift(1, h-1)
            g = 1

            do i=1, h
                id(i) = g
                if (and(rshift(div-1,i-1), 1) == 1) g=g+1
            end do

            allocate(c(w, maxval(id)))
            c(:,:) = 0

            do i=1,h
                do j=1,w
                    read(s(j,i),*) ints
                    c(j,id(i))=c(j,id(i))+ints
                end do 
            end do

            ok=.true.
            do i=1,g
                do j=1,w
                    if(c(j,i) > k) ok = .false.
                end do
            end do

            if (.not. ok) cycle

            allocate(now(g+1))
            num = g-1
            now(:) = 0
            do j=1,w
                if (.not. add(j,g)) then
                    num=num+1
                    now(:)=0
                    emp = add(j,g)
                end if
            end do 

            ans = min(ans,num)
            deallocate(c)
            deallocate(now)
        end do

        print*, ans
    end block

contains
function add(j,g) result(ok)
    integer(8):: i,j,g
    logical:: ok

    do i=1,g
        now(i)=now(i)+c(j,i)
    end do

    ok=.true.
    do i=1,g+1
        if (now(i) > k) ok=.false.
    end do
end function

end program main