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 <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <numeric>
#include <functional>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <string>
#define repd(i,a,b) for (int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repd(i,0,n)
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
#define inf 2000000007
#define mp make_pair
#define pb push_back
typedef long long ll;
using namespace std;
template <typename T>
inline void output(T a, int p) {
    if(p) cout << fixed << setprecision(p)  << a << "\n";
    else cout << a << "\n";
}
// end of template
int dx[9] = {0, 1, -1, 0, 0, 1, 1, -1, -1};
int dy[9] = {0, 0, 0, 1, -1, 1, -1, 1, -1};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    // source code
    int N, M;
    cin >> N >> M;
    vector<vector<int>> A(N, vector<int>(M, 0));
    rep(i, N) rep(j, M){
        char c;
        cin >> c;
        if(i == 0) {
            A[i][j] = 1;
        }
        if(c == 'o'){
            rep(k, 3){
                if(i + dy[k] >= 0 && i + dy[k] < N && j + dx[k] >= 0 && j + dx[k] < M)
                    A[i + dy[k]][j + dx[k]] = 1;
            }
        }
        if(c == 'x'){
            rep(k, 9){
                if(i + dy[k] >= 0 && i + dy[k] < N && j + dx[k] >= 0 && j + dx[k] < M)
                    A[i + dy[k]][j + dx[k]] = 1;
            }
        }
    }
    int ret = 0;
    rep(i, N) rep(j, M){
        if(A[i][j] == 0) ret++;
    }
    
    cout << ret << endl;
    return 0;
}