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
#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
const ll mod=1e9+7;
const int inf=1<<30;

int h,w;
vector<string> g;
int dy[]={0,0,1,-1};
int dx[]={1,-1,0,0};
ll bcnt,wcnt;
bool seen[410][410];
void dfs(int y,int x){
  seen[y][x]=1;
  if(g[y][x]=='.') wcnt++;
  if(g[y][x]=='#') bcnt++;
  rep(dir,4){
    int ny=y+dy[dir];
    int nx=x+dx[dir];
    if(ny<0||ny>=h||nx<0||nx>=w) continue;
    if(seen[ny][nx]) continue;
    if(g[y][x]==g[ny][nx]) continue;
    dfs(ny,nx);
  }
}
int main(){
  cin >> h >> w;
  g.resize(h);
  rep(i,h) cin >> g[i];
  ll ans=0;
  rep(i,h)rep(j,w){
    bcnt=0; wcnt=0;
    dfs(i,j);
    ans+=bcnt*wcnt;
  }
  cout << ans << endl;
}