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>
using namespace std;
using ll=long long;
#define rep(i,a) for(ll i=0;i<a;i++)

class UnionFind{
public:
    vector<ll>data;
    UnionFind(ll n):data(n,-1){}
    bool Union(ll a,ll b){
        a=root(a);b=root(b);
        if(a==b)return 1;
        if(data[a]>data[b])swap(a,b);
        data[a]+=data[b];
        data[b]=a;
        return 0;
    }
    bool Find(ll a,ll b){return root(a)==root(b);}
    ll root(ll a){return data[a]<0?a:root(data[a]);}
};

signed main(){
    ll h,w;
    cin>>h>>w;
    UnionFind uf(h*w);
    vector<string>s(h);
    for(auto&i:s)cin>>i;
    rep(i,h)rep(j,w){
        if(i+1<h&&s[i][j]!=s[i+1][j])uf.Union(i*w+j,(i+1)*w+j); // 下が違う色だったらまとめる
        if(j+1<w&&s[i][j]!=s[i][j+1])uf.Union(i*w+j,i*w+j+1); // 右が違う色だったらまとめる
    }
    vector<pair<ll,ll>>cnt(h*w);
    rep(i,h)rep(j,w){
        if(s[i][j]=='#')cnt[uf.root(i*w+j)].first++; // rootで白と黒を数える
        else cnt[uf.root(i*w+j)].second++;
    }
    ll ans=0;
    rep(i,h*w)ans+=cnt[i].first*cnt[i].second;
    cout<<ans;
}
