Suppose that you are in a campus and have to go for classes day by day. As you may see, when you hurry to a classroom, you surprisingly find that many seats there are already occupied. Today you and your friends went for class, and found out that some of the seats were occupied.

The classroom contains n rows of seats and there are m seats in each row. Then the classroom can be represented as an n × m matrix. The character '.' represents an empty seat, while '*' means that the seat is occupied. You need to find k consecutive empty seats in the same row or column and arrange those seats for you and your friends. Your task is to find the number of ways to arrange the seats. Two ways are considered different if sets of places that students occupy differs.

Input

The first line contains three positive integers n,m,k (1 ≤ n, m, k ≤ 2 000), where n,m represent the sizes of the classroom and k is the number of consecutive seats you need to find.

Each of the next n lines contains m characters '.' or '*'. They form a matrix representing the classroom, '.' denotes an empty seat, and '*' denotes an occupied seat.

Output

A single number, denoting the number of ways to find k empty seats in the same row or column.

Examples

Input

2 3 2
**.
...


Output

3


Input

1 2 2
..


Output

1


Input

3 3 4
.*.
*.*
.*.


Output

0

Note

In the first sample, there are three ways to arrange those seats. You can take the following seats for your arrangement. 

  * (1,3), (2,3) 
  * (2,2), (2,3) 
  * (2,1), (2,2) 

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.*;

public class SeatArrangements {

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static class Pair {
		int diff;
		int count;

		public Pair(int diff, int count) {
			super();
			this.diff = diff;
			this.count = count;
		}

	}

	public static void main(String[] args) {

		FastScanner f = new FastScanner();
		PrintWriter p = new PrintWriter(System.out);
		int r = f.nextInt();
		int c = f.nextInt();
		int k = f.nextInt();
		char mat[][] = new char[r][c];
		int empty = 0;
		for (int i = 0; i < r; i++) {
			String str = f.nextLine();
			for (int j = 0; j < c; j++) {
				mat[i][j] = str.charAt(j);
				if (mat[i][j] == '.')
					empty++;
			}
		}
//		for(int i=0;i<r;i++) {
//			for(int j=0;j<c;j++) {
//				p.print(mat[i][j]);
//			}
//			p.println();
//		}

		if (k == 1) {
			p.println(empty);
			p.close();
			return;
		}

		// traverse in each row;
		int ans = 0;
		for (int i = 0; i < r; i++) {
			int cons = 0;
			for (int j = 0; j < c; j++) {
				if (mat[i][j] == '.')
					cons++;
				else {
					if (cons >= k) {
						ans += cons - (k - 1);
					}
					cons = 0;
				}
			}
			if (cons >= k) {
				ans += cons - (k - 1);
			}
			cons = 0;
		}

		// traverse in each column
		for (int i = 0; i < c; i++) {
			int cons = 0;
			for (int j = 0; j < r; j++) {
				if (mat[j][i] == '.')
					cons++;
				else {
					if (cons >= k) {
						ans += cons - (k - 1);
					}
					cons = 0;
				}
			}
			if (cons >= k) {
				ans += cons - (k - 1);
			}
			cons = 0;
		}

		p.println(ans);
		p.close();

	}

	/// Ceil
	static long ceil(long x, long m) {
		long res = x / m;
		if (x % m != 0) {
			res++;
		}
		return res;
	}

	// ------------------------------------------------------------------------------------------------
	// makes the prefix sum array
	static long[] prefixSum(long arr[], int n) {
		long psum[] = new long[n];
		psum[0] = arr[0];
		for (int i = 1; i < n; i++) {
			psum[i] = psum[i - 1] + arr[i];
		}
		return psum;
	}
	// ------------------------------------------------------------------------------------------------

	// makes the suffix sum array
	static long[] suffixSum(long arr[], int n) {
		long ssum[] = new long[n];
		ssum[n - 1] = arr[n - 1];
		for (int i = n - 2; i >= 0; i--) {
			ssum[i] = ssum[i + 1] + arr[i];
		}
		return ssum;
	}

//------------------------------------------------------------------------------------------
	// BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
	// OVERHEADS
	static long m = (long) (1e9 + 7);

	static long binPower(long a, long n, long m) {
		if (n == 0)
			return 1;
		long res = 1;
		while (n > 0) {
			if ((n & 1) != 0) {
				res *= a;
			}
			a *= a;
			n >>= 1;
		}
		return res;
	}

//-------------------------------------------------------------------------------------------

	// gcd
	static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

//------------------------------------------------------------------------------------------

	// lcm
	static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}

//------------------------------------------------------------------------------------------
	// BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
	// O(LOGn)
	static int setBits(int n) {
		int count = 0;
		while (n > 0) {
			n = n & (n - 1);
			count++;
		}
		return count;
	}
//------------------------------------------------------------------------------------------

//------------------------------------------------------------------------------------------
	// 0 based indexing
	static boolean KthBitSet(int n, int k) {
		int mask = 1;
		mask = mask <<= k;
		if ((mask & n) != 0)
			return true;
		else
			return false;
	}

//------------------------------------------------------------------------------------------

	// EXTENDED EUCLIDEAN THEOREM
	// TO REPRESENT GCD IN TERMS OF A AND B
	// gcd(a,b) = a.x + b.y where x and y are integers

	static long x = -1;
	static long y = -1;

	static long gcdxy(long a, long b) {
		if (b == 0) {
			x = 1;
			y = 0;
			return a;
		} else {
			long d = gcdxy(b, a % b);
			long x1 = y;
			long y1 = x - (a / b) * y;
			x = x1;
			y = y1;

			return d;
		}
	}

//-------------------------------------------------------------------------------------------------F
}
