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.

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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
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vpii = vector<pii>;
template <int mxn>
struct BITQ {
  int n;
  vector<pair<pii, int>> a;
  void clear() {
    a.clear();
    n = 0;
  }
  void push(int op, pii x) {
    a.push_back({x, op});
    ++n;
  }
  int d[mxn];
  void set(int x, int val) {
    for (int i = x; i >= 1; i -= ((i) & -(i))) d[i] += val;
  }
  int get(int x) {
    int re = 0;
    for (int i = x; i < mxn; i += ((i) & -(i))) re += d[i];
    return re;
  }
  int sum() {
    int re = 0;
    sort(a.begin(), a.end());
    for (auto &i : a)
      if (i.second < 0)
        set(-i.first.second, 1);
      else
        re += get(-i.first.second);
    for (auto &i : a)
      if (i.second < 0) set(-i.first.second, -1);
    return re;
  }
};
const int mxn = 3006;
int n, m;
char a[mxn][mxn];
int L[mxn][mxn];
int R[mxn][mxn];
BITQ<mxn> Q;
void EXEC() {
  scanf("%d%d\n", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%s", a[i] + 1);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j)
      L[i][j] = (a[i][j] == 'z' ? L[i][j - 1] + 1 : 0);
    for (int j = m; j >= 1; --j)
      R[i][j] = (a[i][j] == 'z' ? R[i][j + 1] + 1 : 0);
  }
  auto f = [](int x) { return (0 + x - 1) * x / 2; };
  ll ans = 0;
  for (int d = 1 + 1; d <= n + m; ++d) {
    Q.clear();
    for (int i = min(d - 1, n), j = d - i; i >= 1 && j <= m; --i, ++j)
      if (a[i][j] == 'z') {
        int &x = j;
        Q.push(-1, {-(x), -(x + R[i][j] - 1)});
        Q.push(+1, {-(x - L[i][j] + 1), -(x)});
      } else {
        ans += Q.sum() - f(Q.n / 2);
        Q.clear();
      }
    ans += Q.sum() - f(Q.n / 2);
  }
  cout << ans << endl;
}
int main() {
  EXEC();
  return 0;
}
