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>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

typedef int _loop_int;
#define REP(i,n) for(_loop_int i=0;i<(_loop_int)(n);++i)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1;i>=(_loop_int)(a);--i)

#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define DEBUG_VEC(v) cout<<#v<<":";REP(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

#define CHMIN(a,b) a=min((a),(b))
#define CHMAX(a,b) a=max((a),(b))

// mod
const ll MOD = 1000000007ll;
#define FIX(a) ((a)%MOD+MOD)%MOD

// floating
typedef double Real;
const Real EPS = 1e-11;
#define EQ0(x) (abs(x)<EPS)
#define EQ(a,b) (abs(a-b)<EPS)
typedef complex<Real> P;

int n,m;
char mp[125][125];
int ban[125][125];

int main(){
  scanf("%d%d",&n,&m);
  REP(i,n)scanf("%s",mp[i]);
  REP(i,m)ban[0][i]=1;
  REP(i,n)REP(j,m){
    if(mp[i][j]=='o'){
      ban[i][j]=1;
      if(j>0)ban[i][j-1]=1;
      if(j<m-1)ban[i][j+1]=1;
    }
    if(mp[i][j]=='x'){
      FOR(x,-1,2)FOR(y,-1,2){
        if(i+x>=0&&i+x<n&&j+y>=0&&j+y<m)ban[i+x][j+y]=1;
      }
    }
  }
  int ans = 0;
  REP(i,n)REP(j,m)ans+=1-ban[i][j];
  cout<<ans<<endl;
  return 0;
}