A: Taking a Seat-Taking a Seat-

story

Mr. A entered the classroom at the examination site to take an examination. However, Mr. A is a very nervous type. Therefore, depending on the situation, there may be seats that Mr. A does not want to sit in. Therefore, Mr. A decided to find out how many seats he could sit on based on certain conditions.

problem

Mr. A is a nervous type of person. First of all, Mr. A cannot sit in the first row from the front. Because when I sit near the proctor, I get nervous and can't solve the problem. Also, if there is a person on the right or left, Mr. A cannot sit down. This is because the sound of the pencil next to me makes it impossible to concentrate during the test. In addition, you can't sit near a noisy guy (near 8 diagonally up, down, left, and right). This is because my voice echoes in my head during the break and I can't rest. Also, it's natural, but I can't sit in the seat where the students are already sitting.

Is there a place where Mr. A can sit under these conditions? Assuming that M x N seats are lined up in the classroom, let's find the number.

Input format

The first row gives M (number of rows: number of vertical seats) and N (number of columns: number of horizontal seats). From the second line, M lines and a character string of length N are given line by line. The j character of the i-th character string represents the state of the seat in the i-th row and the j-th column. Each letter is either'-',' x'or'o', vacant seats'-', noisy students sitting during breaks'x', other students sitting' Let it be o'.

Constraint

* 0 <M ≤ 100
* 0 <N ≤ 100



Output format

Output the number of seats that Mr. A can sit in one line depending on the conditions. Don't forget the line break at the end.

Input example 1


5 5
--o--
--xo-
--x--
o --- x
--xoo


Output example 1


3

Input example 2


2 6
--oooo
x--o--


Output example 2


1

Input example 3


3 5
-----
-----
-----


Output example 3


Ten

Input example 4


4 6
o-oxoo
oo-ooo
oooxo-
o-ooox


Output example 4


0





Example

Input

5 5
--o--
--xo-
--x--
o---x
--xoo


Output

3
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(n);i++)
#define ALL(A) A.begin(), A.end()

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int dy[8] = {-1,-1, 0, 1, 1, 1, 0,-1}; 
const int dx[8] = { 0, 1, 1, 1, 0,-1,-1,-1};

char s[105][105];

int main()
{
	memset(s, 0, sizeof(s));
	rep (i, 105) rep (j, 105) s[i][j] = '-';
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int M, N; cin >> M >> N;
	rep (i, M) rep (j, N) cin >> s[i+1][j+1];

	int res = 0;
	for (int i = 2; i <= M; ++i){
		for (int j = 1; j <= N; ++j){
			bool ng = false;
			if (s[i][j] != '-') ng |= true;
			if (s[i][j-1] != '-' || s[i][j+1] != '-') ng |= true;
			rep (k, 8)
				if (s[i+dy[k]][j+dx[k]] == 'x') ng |= true;
			if (!ng) ++res;
		} // end for
	} // end for
	cout << res << endl;

	return 0;
}