#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
/*
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("-O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
*/
#define mp make_pair
#define ll long long
#define ld long double
#define pb push_back
#define fast_io ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define fs first
#define sc second
#define getfiles ifstream cin("input.txt"); ofstream cout("output.txt");
#define endl '\n'
#define pii pair<ll, ll>
const ll INF = 2000000005;
const ll BIG_INF = 2000000000000000005;
const ll mod = 1000000007;
const ll P = 31;
const ld PI = 3.141592653589793238462643;
const double eps = 1e-9;
using namespace std;
using namespace __gnu_pbds;
bool valid(ll x, ll y, ll n, ll m) {
return x >= 0 && y >= 0 && x < n && y < m;
}
mt19937 rng(1999999973);
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
const ll N = 2005;
ll n, m, nxt_r[N][N], ans = 0;
string s[N];
struct seg {
ll l, r, mn;
seg(ll l, ll r, ll mn) {
this->l = l;
this->r = r;
this->mn = mn;
}
seg() {
l = -1, r = -1, mn = 0;
}
};
inline ll ariphm(ll l, ll r) {
if (l > r)
return 0;
return (l + r) * (r - l + 1) / 2;
}
signed main() {
fast_io;
cin >> n >> m;
for (ll i = 0; i < n; i++)
cin >> s[i];
for (ll i = 0; i < n; i++) {
nxt_r[i][m - 1] = s[i][m - 1] != '#';
for (ll j = m - 2; j >= 0; j--) {
nxt_r[i][j] = nxt_r[i][j + 1] + 1;
if (s[i][j] == '#')
nxt_r[i][j] = 0;
}
}
for (ll j = 0; j < m; j++) {
vector<seg> st(1, seg());
ll sum1 = 0, sum2 = 0;
for (ll i = 0; i < n; i++) {
while(st.back().mn > nxt_r[i][j]) {
sum1 -= ariphm(st.back().l, st.back().r) * ariphm(1, st.back().mn);
sum2 -= ariphm(1, st.back().mn) * (st.back().r - st.back().l + 1);
st.pop_back();
}
st.pb(seg(st.back().r + 1, i, nxt_r[i][j]));
sum1 += ariphm(st.back().l, st.back().r) * ariphm(1, st.back().mn);
sum2 += ariphm(1, st.back().mn) * (st.back().r - st.back().l + 1);
ans += sum2 * (i + 1) - sum1;
}
}
cout << ans;
return 0;
}
