#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
const int M = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
const int oo = 1000000000;
int R, C, rr[3001][3001], ll[3001][3001], dd[3001][3001], fen[3001], l, r;
vector<int> k;
vector<pair<int, int> > v;
char g[3001][3001];
void add(int i) {
  for (; i < 3001; i += i & -i) ++fen[i];
}
int get(int i) {
  int res = 0;
  for (; i; i -= i & -i) res += fen[i];
  return res;
}
int main() {
  cin >> R >> C;
  for (int i = 1; i <= R; ++i) scanf("%s", g[i] + 1);
  for (int i = 1; i <= R; ++i)
    for (int j = 1; j <= C; ++j)
      if (g[i][j] == 'z')
        ll[i][j] = 1 + ll[i][j - 1], dd[i][j] = 1 + dd[i - 1][j + 1];
  for (int i = 1; i <= R; ++i)
    for (int j = C; j; --j)
      if (g[i][j] == 'z') rr[i][j] = 1 + rr[i][j + 1];
  long long ans = 0;
  for (int a = 1, b = 1; b <= C; a < R ? ++a : ++b) {
    v.clear();
    k.clear();
    k.resize(a + 1);
    for (int i = a, j = b; i > 0 && j <= C; --i, ++j) {
      if (g[i][j] == '.') continue;
      int p = min(rr[i][j] - 1, dd[i][j] - 1);
      v.push_back(pair<int, int>(i - p, i));
      k[i] = min(R, i + ll[i][j] - 1);
    }
    sort((v).begin(), (v).end());
    memset(fen, 0, sizeof(fen));
    int p = 1;
    for (int t = 0; t < v.size(); ++t) {
      int d = v[t].first;
      while (p < d) {
        if (k[p]) ans += get(k[p]) - get(p - 1);
        ++p;
      }
      add(v[t].second);
    }
    while (p < k.size()) {
      if (k[p]) ans += get(k[p]) - get(p - 1);
      ++p;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

A tourist wants to visit country Zeydabad for Zbazi (a local game in Zeydabad).

The country Zeydabad is a rectangular table consisting of n rows and m columns. Each cell on the country is either 'z' or '.'.

The tourist knows this country is named Zeydabad because there are lots of ''Z-pattern"s in the country. A ''Z-pattern" is a square which anti-diagonal is completely filled with 'z' and its upper and lower rows are also completely filled with 'z'. All other cells of a square can be arbitrary.

<image>

Note that a ''Z-pattern" can consist of only one cell (see the examples).

So he wants to count the number of ''Z-pattern"s in the country (a necessary skill for Zbazi).

Now your task is to help tourist with counting number of ''Z-pattern"s.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 3000) — the number of rows and columns respectively.

Each of the next n lines contains m characters 'z' or '.' — the description of Zeydabad.

Output

Print the only integer a — the number of ''Z-pattern"s in Zeydabad.

Examples

Input

4 4
zzzz
zzz.
.z..
zzzz


Output

16


Input

1 4
z.z.


Output

2


Input

2 2
zz
zz


Output

5

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
const int M = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
const int oo = 1000000000;
int R, C, rr[3001][3001], ll[3001][3001], dd[3001][3001], fen[3001], l, r;
vector<int> k;
vector<pair<int, int> > v;
char g[3001][3001];
void add(int i) {
  for (; i < 3001; i += i & -i) ++fen[i];
}
int get(int i) {
  int res = 0;
  for (; i; i -= i & -i) res += fen[i];
  return res;
}
int main() {
  cin >> R >> C;
  for (int i = 1; i <= R; ++i) scanf("%s", g[i] + 1);
  for (int i = 1; i <= R; ++i)
    for (int j = 1; j <= C; ++j)
      if (g[i][j] == 'z')
        ll[i][j] = 1 + ll[i][j - 1], dd[i][j] = 1 + dd[i - 1][j + 1];
  for (int i = 1; i <= R; ++i)
    for (int j = C; j; --j)
      if (g[i][j] == 'z') rr[i][j] = 1 + rr[i][j + 1];
  long long ans = 0;
  for (int a = 1, b = 1; b <= C; a < R ? ++a : ++b) {
    v.clear();
    k.clear();
    k.resize(a + 1);
    for (int i = a, j = b; i > 0 && j <= C; --i, ++j) {
      if (g[i][j] == '.') continue;
      int p = min(rr[i][j] - 1, dd[i][j] - 1);
      v.push_back(pair<int, int>(i - p, i));
      k[i] = min(R, i + ll[i][j] - 1);
    }
    sort((v).begin(), (v).end());
    memset(fen, 0, sizeof(fen));
    int p = 1;
    for (int t = 0; t < v.size(); ++t) {
      int d = v[t].first;
      while (p < d) {
        if (k[p]) ans += get(k[p]) - get(p - 1);
        ++p;
      }
      add(v[t].second);
    }
    while (p < k.size()) {
      if (k[p]) ans += get(k[p]) - get(p - 1);
      ++p;
    }
  }
  cout << ans << endl;
  return 0;
}

```