D. Timetable

http://codeforces.com/problemset/problem/946/D

Ivan is a student at Berland State University (BSU). There are n days in Berland week, and each of these days Ivan might have some classes at the university.

There are m working hours during each Berland day, and each lesson at the university lasts exactly one hour. If at some day Ivan's first lesson is during i-th hour, and last lesson is during j-th hour, then he spends j - i + 1 hours in the university during this day. If there are no lessons during some day, then Ivan stays at home and therefore spends 0 hours in the university.

Ivan doesn't like to spend a lot of time in the university, so he has decided to skip some lessons. He cannot skip more than k lessons during the week. After deciding which lessons he should skip and which he should attend, every day Ivan will enter the university right before the start of the first lesson he does not skip, and leave it after the end of the last lesson he decides to attend. If Ivan skips all lessons during some day, he doesn't go to the university that day at all.

Given n, m, k and Ivan's timetable, can you determine the minimum number of hours he has to spend in the university during one week, if he cannot skip more than k lessons?

Input

The first line contains three integers n, m and k (1 ≤ n, m ≤ 500, 0 ≤ k ≤ 500) — the number of days in the Berland week, the number of working hours during each day, and the number of lessons Ivan can skip, respectively.

Then n lines follow, i-th line containing a binary string of m characters. If j-th character in i-th line is 1, then Ivan has a lesson on i-th day during j-th hour (if it is 0, there is no such lesson).

Output

Print the minimum number of hours Ivan has to spend in the university during the week if he skips not more than k lessons.

Examples

input

Copy

2 5 1
01001
10110

output

5

input

Copy

2 5 0
01001
10110

output

8

Note

In the first example Ivan can skip any of two lessons during the first day, so he spends 1 hour during the first day and 4 hours during the second day.

In the second example Ivan can't skip any lessons, so he spends 4 hours every day.

// 去吧！皮卡丘! 把AC带回来！
//      へ　　　　　／|
// 　　/＼7　　　 ∠＿/
// 　 /　│　　 ／　／
// 　│　Z ＿,＜　／　　 /`ヽ
// 　│　　　　　ヽ　　 /　　〉
// 　Y　　　　　`　 /　　/
// 　ｲ●　､　●　　⊂⊃〈　　/
// 　()　 へ　　　　|　＼〈
// 　　>ｰ ､_　 ィ　 │ ／／
// 　 / へ　　 /　ﾉ＜| ＼＼
// 　 ヽ_ﾉ　　(_／　 │／／
// 　　 7　　　　　　　|／
// 　　 ＞―r￣￣`ｰ―＿
//**************************************
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 2147483647
const ll INF = 0x3f3f3f3f3f3f3f3fll;
#define ri register int
template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
template <class T> inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T> inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
#define pi acos(-1)
#define mem(x, y) memset(x, y, sizeof(x));
#define For(i, a, b) for (int i = a; i <= b; i++)
#define FFor(i, a, b) for (int i = a; i >= b; i--)
#define bug printf("***********
");
#define mp make_pair
#define pb push_back
const int maxn = 3e5 + 10;
// name*******************************
int n, m;
int a[505][505];
int K;
int kk[505][505]；//kk[i][j]记录第i行删掉j个的最小代价，为后面dp服务
int dp[505][505];//dp[i][j]：前i行，共删掉j个所获最下代价
int cnt[505];//记录每组的1的个数
int sum=0;  //sum这里记录下1的总个数，若K大于sum，直接输出0
// function******************************

//***************************************
int main() {
  // ios::sync_with_stdio(0); cin.tie(0);
  // freopen("test.txt", "r", stdin);
  //  freopen("outout.txt","w",stdout);
  cin >> n >> m >> K;
  me(dp, 127);
  me(kk, 127);
  For(i, 1, n) {
    For(j, 1, m) {
      int x;
      scanf("%1d", &x);//限宽输入
      if (x) {
        a[i][++cnt[i]] = j;
      }
    }
    //选定j,k为起始终止点
    For(j, 1, cnt[i]) {
      For(k, j, cnt[i]) {
        kk[i][(j - 1) + cnt[i] - k] =
            min(kk[i][(j - 1) + cnt[i] - k], a[i][k] - a[i][j] + 1);
      }
    }
    kk[i][cnt[i]] = 0;
    sum+=cnt[i];
  }
if(K>=sum){
  cout<<0;
  return 0;
}
  For(i, 0, min(K, cnt[1])) {
    dp[1][i] = kk[1][i];
  }
  For(i, 2, n) {
    For(j, 0, K) {
      For(k, 0, min(j,cnt[i])) {
        dp[i][j] = min(dp[i][j], dp[i - 1][j - k] + kk[i][k]);
      }
    }
  }
  cout << dp[n][K];

  return 0;
}