Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

• Line 1: Two space-separated integers: N and M

• Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

• Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12

W........WW.

.WWW.....WWW

....WW...WW.

.........WW.

.........W..

..W......W..

.W.W.....WW.

W.W.W.....W.

.W.W......W.

..W.......W.

Sample Output

3

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

搜索水题，只要往四周一直搜，搜过的改变值就可以，能搜几次就是几个；

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include <iomanip>

#include<cmath>

#include<float.h>

#include<string.h>

#include<algorithm>

#define sf scanf

#define pf printf

#define pb push_back

#define mm(x,b) memset((x),(b),sizeof(x))

#include<vector>

#include<map>

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=a;i>=n;i--)

typedef long long ll;

typedef long double ld;

typedef double db;

const ll mod=1e12+100;

const db e=exp(1);

using namespace std;

const double pi=acos(-1.0);

char a[105][105];

bool dfs(int x,int y)

{

//	cout<<x<<" "<<y<<endl;

	if(a[x][y]!='W') return false;

	a[x][y]='a';//把经过的位置改变 

	if(a[x-1][y-1]=='W')//左上

		dfs(x-1,y-1);

	if(a[x-1][y+1]=='W')//右上

		dfs(x-1,y+1);

	if(a[x+1][y-1]=='W')//左下

		dfs(x+1,y-1);

	if(a[x+1][y+1]=='W')//右下

		dfs(x+1,y+1);

	if(a[x-1][y]=='W')//上

		dfs(x-1,y);

	if(a[x][y-1]=='W')//左

		dfs(x,y-1);

	if(a[x][y+1]=='W')//右

		dfs(x,y+1);

	if(a[x+1][y]=='W')//下

		dfs(x+1,y);

		return true;

}

int solve(int n,int m)

{

	int sum=0;

	rep(i,1,n+1)

	{

		rep(j,1,m+1)

		if(dfs(i,j))

		sum++;

	}

	return sum;

}

int main()

{

		int n,m;

		sf("%d%d%d%d",&n,&m);

		mm(a,'.');

		rep(i,1,n+1)

		{

			sf("%s",&a[i][1]);

			//pf("1%s\n",&d[i]);

			a[i][m+1]='.';

		}

		pf("%d\n",solve(n,m));

}