There is a grid with H rows and W columns, where each square is painted black or white.

You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`.

Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition:

* There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...

Constraints

* 1 \leq H, W \leq 400
* |S_i| = W (1 \leq i \leq H)
* For each i (1 \leq i \leq H), the string S_i consists of characters `#` and `.`.

Input

Input is given from Standard Input in the following format:


H W
S_1
S_2
:
S_H


Output

Print the answer.

Examples

Input

3 3
.#.
..#
#..


Output

10


Input

3 3
.#.
..#
..


Output

10


Input

2 4
....
....


Output

0


Input

4 3


...


Output

6
#include<bits/stdc++.h>
typedef long long ll;
#define INF 100000000
#define MOD 100000007
int dx[4]={1,0,-1,0},dy[4]={0,-1,0,1};
using namespace std;
vector<string>s;
bool visit[405][405];
int h,w;
ll nb=0,nw=0;
void dfs(int y,int x){
    visit[y][x]=true;
    if(s[y][x]=='#')nb++;
    else nw++;
    for(int i=0;i<4;i++){
        int nextx=x+dx[i];
        int nexty=y+dy[i];
        if(nextx<0 or nextx>=w or nexty<0 or nexty>=h){
            continue;
        }
        if(visit[nexty][nextx])continue;
        if(s[nexty][nextx]==s[y][x])continue;
        dfs(nexty,nextx);
    }
}
int main(void){
    cin>>h>>w;
    s.resize(h);
    for(int i=0;i<h;i++){
        cin>>s[i];
    }
    ll ans=0;
    for(int y=0;y<h;y++){
        for(int x=0;x<w;x++){
            if(s[y][x]=='.')continue;
            if(visit[y][x])continue;
            nb=0;
            nw=0;
            dfs(y,x);
            ans+=nb*nw;
        }
    }
    cout<<ans<<endl;
   return 0;
}