#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
using namespace std; bool MEM;
using ll=long long; using ld=long double;
using pii=pair<int,int>; using pll=pair<ll,ll>;
const int I=1e9;
const ll J=1e18,N=1007,xx[8]={-1,0,1,0,-1,-1,1,1},yy[8]={0,-1,0,1,-1,1,1,-1};
const ld B=2;
ll n,m,vis[N][N],has[N][N],can[N][N],ans;
char str[N][N];
vector<pll> v;
void dfs(ll x,ll y) {
	vis[x][y]=1,v.pb({x,y});
	for (ll i=0;i<8;i++) {
		ll nx=x+xx[i],ny=y+yy[i];
		if (str[nx][ny]=='#'&&!vis[nx][ny]) dfs(nx,ny);
	}
}
void dfs1(ll x,ll y) {
	has[x][y]=1;
	for (ll i=0;i<8;i++) {
		ll nx=x+xx[i],ny=y+yy[i];
		if (can[nx][ny]&&!has[nx][ny]) dfs1(nx,ny);
	}
}
pll operator - (pll x,pll y) { return {x.fi-y.fi,x.se-y.se}; }
ll operator * (pll x,pll y) { return x.fi*y.se-x.se*y.fi; }
ll operator & (pll x,pll y) { return x.fi*y.fi+x.se*y.se; }
ld len(pll x) { return sqrt((ld)(x.fi*x.fi+x.se*x.se)); }
ld dis(pll a,pll b,pll c) { return (ld)abs((c-a)*(b-a))/len(b-a); }
mt19937_64 dnr(9);
ll rnd(ll l,ll r) { return dnr()%(r-l+1)+l; }
void solve(ll sx,ll sy) {
	v.clear(),dfs(sx,sy);
	sort(v.begin(),v.end());
	ll mx=0;
	for (ll o=0;o<30&&mx<=1;o++) {
		pll s=v[rnd(0,(ll)v.size()-1)],t;
		while (t=v[rnd(0,(ll)v.size()-1)],s==t);
		for (pll i:v) {
			ll d=(t-s)&(i-s);
			has[i.fi][i.se]=0,can[i.fi][i.se]=dis(s,t,i)<=B&&0<=d&&d<=len(t-s)*len(t-s);
		}
		ll sum=0;
		for (pll i:v) if (can[i.fi][i.se]&&!has[i.fi][i.se]) dfs1(i.fi,i.se),sum++;
		mx=max(mx,sum);
	}
	ans+=mx<=1?1:5;
}
void mian() {
	scanf("%lld%lld",&n,&m);
	for (ll i=1;i<=n;i++) scanf("%s",str[i]+1);
	for (ll i=1;i<=n;i++) for (ll j=1;j<=m;j++) if (str[i][j]=='#'&&!vis[i][j]) solve(i,j);
	cout<<ans-5;
}
bool ORY; int main() {
	// while (1)
	// int t; for (scanf("%d",&t);t--;)
	mian();
	cerr<<"\n"<<abs(&MEM-&ORY)/1048576<<"MB";
	return 0;
}