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 <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <cmath>
#include <numeric>
#include <iomanip>
#include <stack>
#include <complex>
#include <functional>
#include <tuple>

using namespace std;

#define Rep(i,a,b) for(ll i = a; i < b; ++i)
#define rep(i,b) Rep(i,0,b)
#define allof(a) (a).begin(), (a).end()

using ll = long long;

constexpr int inf = 1e9 + 7;
constexpr ll infll = 1ll << 60ll;
constexpr ll mod = 1e9 + 7;
// 0~3までは右下左上 4~7までは斜め
constexpr int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
constexpr int dy[] = { 0, 1, 0, -1, -1, 1, 1, -1 };

template<typename T> bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<typename T> bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int main() {

    int m, n;
    cin >> m >> n;
    vector<vector<char>> s(m, vector<char>(n));
    rep(i, m) rep(j, n) cin >> s[i][j];

    int ans = 0;
    Rep(y, 1, m) {
        rep(x, n) {
            if (s[y][x] != '-') continue;
            bool can = true;
            rep(i, 8) {
                int ty = y + dy[i];
                int tx = x + dx[i];

                if (ty < 0 || ty >= m) continue;
                if (tx < 0 || tx >= n) continue;

                if (s[ty][tx] == 'x') {
                    can = false;
                    break;
                }
            }
            // 左右
            int tx = x - 1;
            if (tx >= 0) can &= (s[y][tx] != 'o');
            tx = x + 1;
            if (tx < n) can &= (s[y][tx] != 'o');

            if (can) ++ans;
        }
    }

    cout << ans << endl;

    return 0;
}

