#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int a[N][N]; // 原始数组
int s[N][N]; // 二维前缀和数组

int get(int x1, int y1, int x2, int y2) {
  return s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
}
/*
4 5 5
00000
01111
00011
00011

参考答案：6
*/
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  char c;
  // 输入原始数组
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      cin >> c, a[i][j] = c - '0';
  // 构建二维前缀和数组
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];

  // 遍历每个可能的矩形，计算此矩形的二维前缀和，要求大于等k,求得最小的长乘以宽的值，也就是最小矩形大小
  int ans = 1e9;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      for (int x = i; x <= n; x++)
        for (int y = j; y <= m; y++)
          if (get(i, j, x, y) >= k)
            ans = min(ans, (x - i + 1) * (y - j + 1));

  if(ans==1e9) ans=0;
  cout << ans << endl;
  return 0;
}