Once upon a time there were several little pigs and several wolves on a two-dimensional grid of size n × m. Each cell in this grid was either empty, containing one little pig, or containing one wolf.

A little pig and a wolf are adjacent if the cells that they are located at share a side. The little pigs are afraid of wolves, so there will be at most one wolf adjacent to each little pig. But each wolf may be adjacent to any number of little pigs.

They have been living peacefully for several years. But today the wolves got hungry. One by one, each wolf will choose one of the little pigs adjacent to it (if any), and eats the poor little pig. This process is not repeated. That is, each wolf will get to eat at most one little pig. Once a little pig gets eaten, it disappears and cannot be eaten by any other wolf.

What is the maximum number of little pigs that may be eaten by the wolves?

Input

The first line contains integers n and m (1 ≤ n, m ≤ 10) which denotes the number of rows and columns in our two-dimensional grid, respectively. Then follow n lines containing m characters each — that is the grid description. "." means that this cell is empty. "P" means that this cell contains a little pig. "W" means that this cell contains a wolf. 

It is guaranteed that there will be at most one wolf adjacent to any little pig.

Output

Print a single number — the maximal number of little pigs that may be eaten by the wolves.

Examples

Input

2 3
PPW
W.P


Output

2


Input

3 3
P.W
.P.
W.P


Output

0

Note

In the first example, one possible scenario in which two little pigs get eaten by the wolves is as follows. 

<image>
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.StringTokenizer;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.Math.floor;
import static java.lang.Math.ceil;
import static java.lang.Math.pow;
import static java.lang.Math.log;
import static java.lang.Math.sqrt;

/* ------------ Theorems and Lemmas ------------ */

// Fermat's Little Theorem
// (1) ((a ** p) - a) % p = 0; if 'p' is prime
// (2) Further, if 'a % p != 0' (a and b should be coprime i.e. gcd(a, b) = 1), then ((a ** (p - 1)) - 1) % p = 0

/* -------------- End of theorems -------------- */

public class Solution {

	// Digit DP
	private static int digit;
	private static int times;
	private static StringBuilder number;
	private static int[][][] digitDP;
	// End

	protected static final String[] FACTS = new String[] { "", "", "2", "3", "322", "5", "53", "7", "7222", "7332" };

	public static void main(String[] args) {
		FastReader reader = new FastReader();
		int t = 1;
		// t = reader.nextInt();
		StringBuilder ans = new StringBuilder("");
		while (t-- > 0) {
			int n = reader.nextInt();
			int m = reader.nextInt();
			char[][] matrix = new char[n][m];
			for (char[] row : matrix)
				System.arraycopy(reader.next().toCharArray(), 0, row, 0, m);
			int cnt = 0;
			for (int i = 0; i < n; ++i)
				for (int j = 0; j < m; ++j)
					if (matrix[i][j] == 'W') {
						if (i - 1 >= 0 && matrix[i - 1][j] == 'P') {
							++cnt;
							matrix[i - 1][j] = 'E';
						} else if (j - 1 >= 0 && matrix[i][j - 1] == 'P') {
							++cnt;
							matrix[i][j - 1] = 'E';
						} else if (j + 1 < m && matrix[i][j + 1] == 'P') {
							++cnt;
							matrix[i][j + 1] = 'E';
						} else if (i + 1 < n && matrix[i + 1][j] == 'P') {
							++cnt;
							matrix[i + 1][j] = 'E';
						}
					}
			ans.append(cnt + "\n");
		}
		reader.close();
		System.out.println(ans.toString().trim());
	}

	public static boolean prime(int n) {
		if (n <= 1)
			return false;
		if (n == 2)
			return true;
		if ((n & 1) == 0)
			return false;
		int val = (int) sqrt((double) n);
		for (int i = 3; i <= val; ++i)
			if (n % i == 0)
				return false;
		return true;
	}

	public static int ddp(int n) {
		number = new StringBuilder(String.valueOf(n));
		digitDP = new int[12][12][2];
		for (int[][] a : digitDP)
			for (int[] b : a)
				Arrays.fill(b, -1);
		return call(0, 0, 0);
	}

	// Digit DP
	public static int call(int pos, int cnt, int f) {
		if (cnt > times)
			return 0;
		if (pos == number.length())
			return cnt == times ? 1 : 0;
		if (digitDP[pos][cnt][f] != -1)
			return digitDP[pos][cnt][f];
		int res = 0;
		int lmt = f == 0 ? number.charAt(pos) - '0' : 9;
		for (int dgt = 0; dgt <= lmt; ++dgt)
			if ((dgt == digit ? 1 + cnt : cnt) <= times)
				res += call(1 + pos, dgt == digit ? 1 + cnt : cnt, f == 0 && dgt < lmt ? 1 : f);
		digitDP[pos][cnt][f] = res;
		return res;
	}

	// Binary Exponentiation
	public static int binpow(int a, int n, int mod) {
		// Iterative approach
		a %= mod;
		int result = 1;
		while (n > 0) {
			if ((n & 1) != 0)
				result = result * a % mod;
			a = a * a % mod;
			n >>= 1;
		}
		return result;

		// Recursive approach
		/*
		 * if (n == 0l) return 1l; long res = binpow(a, n / 2, mod); if ((n & 1l) != 0l)
		 * return ((res % mod) * (res % mod) * (a % mod)) % mod; else return ((res %
		 * mod) * (res % mod)) % mod;
		 */
	}

	// GCD
	public static int gcd(int a, int b) {
		while (b > 0) {
			a %= b;
			int t = a;
			a = b;
			b = t;
		}
		return a;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

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

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

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

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

		float nextFloat() {
			return Float.parseFloat(next());
		}

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

		int[] readArray(int size) {
			int[] ar = new int[size];
			for (int i = 0; i < size; ++i)
				ar[i] = nextInt();
			return ar;
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
	}
}