Problem: How many stars are there in the sky? A young programmer Polycarpus can't get this question out of his head! He took a photo of the starry sky using his digital camera and now he analyzes the resulting monochrome digital picture. The picture is represented by a rectangular matrix consisting of n lines each containing m characters. A character equals '1', if the corresponding photo pixel is white and '0', if it is black.

Polycarpus thinks that he has found a star on the photo if he finds a white pixel surrounded by four side-neighboring pixels that are also white: 
    
    
      
     1   
    111  
     1   
    

a star on the photo

Polycarpus whats to cut out a rectangular area from the photo and give his mom as a present. This area should contain no less than k stars. The stars can intersect, have shared white pixels on the photo. The boy will cut out the rectangular area so that its borders will be parallel to the sides of the photo and the cuts will go straight between the pixel borders.

Now Polycarpus keeps wondering how many ways there are to cut an area out of the photo so that it met the conditions given above. Help Polycarpus find this number.

Input

The first line of the input data contains three integers n, m and k (1 ≤ n, m ≤ 500;1 ≤ k ≤ nm). Then follow n lines, containing the description of the given photo as a sequence of lines. Each line contains m characters '0' or '1'.

Output

Print the required number of areas on the given photo.

Examples

Input

4 6 2
111000
111100
011011
000111


Output

6


Input

5 5 4
11111
11111
11111
11111
11111


Output

9

Note

We'll number the rows and columns below starting from 1, the coordinates (p, q) will denote a cell in row p, column q.

In the first sample Polycarpus should cut out any area containing a rectangle whose opposite corners lie in cells (1, 1) and (3, 4). Only rectangles with opposite corners in (1, 1) and (x, y), where x ≥ 3 and y ≥ 4 fit the conditions.

In the second sample any rectangle whose each side is no less than four, will do. The possible rectangle sizes are 4 × 4, 4 × 5, 5 × 4 and 5 × 5. Such figures can be cut in 4 ways, 2 ways, 2 ways and 1 way correspondingly.

Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use cin, cout streams or the %I64d specificator.

Solution (in JAVA):

Solution:
import static java.util.Arrays.fill;

import java.io.*;
import java.util.*;

public class F {

	private static void solve() throws IOException {
		// stress();
		int n = nextInt(), m = nextInt();
		int k = nextInt();
		boolean[][] b = new boolean[n][m];
		for (int i = 0; i < n; i++) {
			String s = nextToken();
			if (s.length() != m) {
				throw new AssertionError();
			}
			for (int j = 0; j < m; j++) {
				b[i][j] = s.charAt(j) == '1';
			}
		}
		out.println(get(b, k));
	}

	static long get(boolean[][] b, int k) {
		if (k <= 0) {
			throw new AssertionError();
		}
		int n = b.length, m = b[0].length;
		if (n < 3 || m < 3) {
			return 0;
		}
		boolean[][] star = star(b);
		n -= 2;
		m -= 2;
		if (k > n * m) {
			return 0;
		}
		int[] cnt = new int[m];
		long res = 0;
		for (int top = 0; top < n; top++) {
			fill(cnt, 0);
			for (int bottom = top + 1; bottom <= n; bottom++) {
				boolean[] cur = star[bottom - 1];
				for (int i = 0; i < m; i++) {
					if (cur[i])
						++cnt[i];
				}
				res += ways(cnt, k);
			}
		}
		return res;
	}

	static long getStupid(boolean[][] b, int k) {
		int res = 0;
		int n = b.length, m = b[0].length;
		for (int i = 0; i < n; i++) {
			for (int i1 = i + 1; i1 <= n; i1++) {
				for (int j = 0; j < m; j++) {
					for (int j1 = j + 1; j1 <= m; j1++) {
						boolean[][] b1 = new boolean[i1 - i][j1 - j];
						if (b1.length < 3 || b1[0].length < 3) {
							continue;
						}
						for (int l = 0; l < b1.length; l++) {
							for (int l2 = 0; l2 < b1[0].length; l2++) {
								b1[l][l2] = b[i + l][j + l2];
							}
						}
						boolean[][] star = star(b1);
						int cnt = 0;
						for (boolean[] star1 : star) {
							for (boolean star2 : star1) {
								if (star2) {
									cnt++;
								}
							}
						}
						if (cnt >= k) {
							res++;
						}
					}
				}
			}
		}
		return res;
	}

	static Random rng = new Random();

	static void stress() {
		while (true) {
			int n = rng.nextInt(5) + 3;
			int m = rng.nextInt(5) + 3;
			boolean[][] b = new boolean[n][m];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < m; j++) {
					b[i][j] = rng.nextBoolean();
				}
			}
			int k = rng.nextInt(n * m) + 1;
			long got = get(b, k);
			long stup = getStupid(b, k);
			if (got != stup) {
				System.err.println(n + " " + m + " " + k);
				for (int i = 0; i < n; i++) {
					for (int j = 0; j < m; j++) {
						System.err.print(b[i][j] ? '1' : '0');
					}
					System.err.println();
				}
				System.err.println("got " + got);
				System.err.println("stup " + stup);
				throw new AssertionError();
			}
		}
	}

	static int ways(int[] cnt, int k) {
		int ret = 0;
		int n = cnt.length;
		int i = 0, j = 0;
		int sum = 0;
		for (; i < n; i++) {
			if (i > 0) {
				sum -= cnt[i - 1];
			}
			while (j < n && sum < k) {
				sum += cnt[j++];
			}
			if (sum >= k) {
				ret += n - j + 1;
			}
		}
		return ret;
	}

	static boolean[][] star(boolean[][] b) {
		int n = b.length, m = b[0].length;
		boolean[][] star = new boolean[n - 2][m - 2];
		for (int i = 1; i < n - 1; i++) {
			for (int j = 1; j < m - 1; j++) {
				star[i - 1][j - 1] = b[i][j] && b[i - 1][j] && b[i][j - 1] && b[i + 1][j]
						&& b[i][j + 1];
			}
		}
		return star;
	}

	public static void main(String[] args) {
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			System.exit(239);
		}
	}

	static BufferedReader br;
	static StringTokenizer st;
	static PrintWriter out;

	static String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null) {
				return null;
			}
			st = new StringTokenizer(line);
		}
		return st.nextToken();
	}

	static int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	static long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}