# Uva 572 - Oil Deposits

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# 題目大意
給定一個n*m，由@和*形成的圖，求@形成的連通塊個數。連通是指上、右上、右、右下、下、左下、左、左上這八個方向的方格相同

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# 題解
BFS

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#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define pf push_front
#define ft first
#define sec second
#define pr pair<int,int>
#define ISCC ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std; 
int t ,n ,m ,ans ,pos ,vis[105][105] ,dir[][2] = {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};
string g[105] ,s;
void sol(int x ,int y)
{
	queue<pr> que;
	que.push({x,y});
	vis[x][y] = 1;
	while(!que.empty())
	{
		pr now = que.front(); que.pop();
		vis[now.ft][now.sec] = 1;
		for(int i=0 ;i<8 ;i++)
		{
			int gx = now.ft + dir[i][0] ,gy = now.sec + dir[i][1];
			if(gx<1 || gy<1 || gx>n || gy>m) continue;
			if(g[gx][gy]!='@' || vis[gx][gy]) continue;
			que.push({gx ,gy});
		}
	}
}
int main()
{
	ISCC;
	while(cin >> n >> m && n && m)
	{
		ans = 0;
		memset(vis,0,sizeof(vis));
		for(int i=1 ;i<=n ;i++) cin >> s ,g[i] = ' ' + s;
		for(int i=1 ;i<=n ;i++)
			for(int j=1 ;j<=m ;j++) if(g[i][j]=='@' && !vis[i][j])
			sol(i ,j) ,ans++;
		cout << ans << '\n';
	}
	return 0;
}