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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
#define ALL(c) (c).begin(), (c).end()
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())

using namespace std;
using ll = long long;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline void dump(vector<T> v) { for (auto& x : v) cerr << x << " "; cerr << endl; }
template<class T> inline void dump(vector<pair<T, T>> v) { for (auto& p : v) cerr << p.first << " " << p.second << endl; }
template<class T> inline void dump(vector<vector<T>> vv) { for (auto& v : vv) {for (auto& x : v) cerr << x << " "; cerr << endl;} }

constexpr int INF = 1001001001;
constexpr long long INFL = (1LL<<60);
constexpr double eps = (1e-9);

// v を縦方向にカットしていった時の最小のカットの回数
// ただし，全部の切れ目でカットしても条件を満たさない場合は -1 を返す
int solve(const vector<vector<int>>& v, const int& h, const int& w, const int& k) {
  vector<int> r(v.size(), 0);
  int ans = 0;
  for (int j = 0; j < w ; j++) {
    bool cut = false;  
    rep(i, v.size()) {
      if (v[i][j] > k) return -1;
      r[i] += v[i][j];
      if (r[i] > k) {
        cut = true;
        break;
      }
    }
    if (cut) {
      ans++;
      rep(i, v.size()) {
        r[i] = v[i][j];
      }
    }
  }
  return ans;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int h, w, k;
  cin >> h >> w >> k;
  vector<string> s(h);
  rep(i, h) cin >> s[i];

  int ans = INF;

  for (int bit = 0; bit < (1 << (h-1)); bit++) {
    bitset<32> bs(bit);
    vector<vector<int>> v(bs.count() + 1, vector<int>(w, 0));
    int idx = 0;
    for (int i = 0; i < h; i++) {
      rep(j, w) if (s[i][j] == '1') v[idx][j]++;
      if ((bit >> i) & 1) idx++;
    }

    // ここまでOK
    int ret = solve(v, h, w, k);
    if (ret != -1) chmin(ans, ret + int(bs.count()));
  }

  cout << ans << endl;

  return 0;
}