Problem: You take part in the testing of new weapon. For the testing a polygon was created. The polygon is a rectangular field n × m in size, divided into unit squares 1 × 1 in size. The polygon contains k objects, each of which is a rectangle with sides, parallel to the polygon sides and entirely occupying several unit squares. The objects don't intersect and don't touch each other.

The principle according to which the weapon works is highly secret. You only know that one can use it to strike any rectangular area whose area is not equal to zero with sides, parallel to the sides of the polygon. The area must completely cover some of the unit squares into which the polygon is divided and it must not touch the other squares. Of course the area mustn't cross the polygon border. Your task is as follows: you should hit no less than one and no more than three rectangular objects. Each object must either lay completely inside the area (in that case it is considered to be hit), or lay completely outside the area.

Find the number of ways of hitting.

Input

The first line has three integers n, m и k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 90) — the sizes of the polygon and the number of objects on it respectively. Next n lines contain m symbols each and describe the polygon. The symbol "*" stands for a square occupied an object, whereas the symbol "." stands for an empty space. The symbols "*" form exactly k rectangular connected areas that meet the requirements of the task.

Output

Output a single number — the number of different ways to hit a target.

Examples

Input

3 3 3
*.*
...
*..


Output

21


Input

4 5 4
.*.**
...**
**...
...**


Output

38


Input

2 2 1
.*
..


Output

4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
struct rect {
  int i1, j1, i2, j2;
};
int a, b, c, d, i, j, n, m, k;
long long ans;
char mas[1111][1111];
rect ob[111];
int beg[222];
inline bool cmp(const rect &a, const rect &b) {
  if (a.j1 != b.j1)
    return a.j1 < b.j1;
  else if (a.j2 != b.j2)
    return a.j2 < b.j2;
  else if (a.i1 != b.i1)
    return a.i1 < b.i1;
  else
    return a.i2 < b.i2;
}
vector<pair<pair<int, int>, int> > cur(0), cur2(0);
vector<pair<int, int> > good;
vector<int> cs(0);
inline int sum(vector<int> &a) {
  int ans = 0;
  for (int _n(((int)((a).size()) - 1)), i(1); i <= _n; i++) ans += a[i];
  return ans;
}
inline long long calc(int l, int r) {
  cur.clear();
  for (int _n((k)-1), i(0); i <= _n; i++) {
    if (ob[i].i2 < l || ob[i].i1 >= r) continue;
    if (ob[i].i1 >= l && ob[i].i2 < r) {
      cur.push_back(make_pair(make_pair(ob[i].j1, ob[i].j2), 1));
    } else {
      cur.push_back(make_pair(make_pair(ob[i].j1, ob[i].j2), 100));
    }
  }
  if (!(int)((cur).size())) return 0;
  sort((cur).begin(), (cur).end());
  cur2.clear();
  int a = 0;
  while (a < (int)((cur).size())) {
    int b = cur[a].first.second, c = a, d = 0;
    while (c < (int)((cur).size())) {
      if (cur[c].first.first <= b) {
        b = max(b, cur[c].first.second);
        d += cur[c].second;
        c++;
      } else
        break;
    }
    cur2.push_back(make_pair(make_pair(cur[a].first.first, b), d));
    a = c;
  }
  cur = cur2;
  long long ans = 0;
  good.clear();
  cs.clear();
  good.push_back(make_pair(0, cur[0].first.first));
  cs.push_back(0);
  for (int _n(((int)((cur).size())) - 1), i(0); i <= _n; i++) {
    if (cur[i].second > 3) {
      good.clear();
      cs.clear();
      int fr = 0;
      if (i < (int)((cur).size()) - 1) {
        fr = cur[i + 1].first.first - cur[i].first.second;
      } else {
        fr = m - cur[i].first.second;
      }
      pair<int, int> rr = make_pair(0, 0);
      rr.first = cur[i].first.second + 1;
      if (i < (int)((cur).size()) - 1)
        rr.second = cur[i + 1].first.first;
      else
        rr.second = m;
      good.push_back(rr);
      cs.push_back(cur[i].second);
    } else {
      int fr = 0;
      if (i < (int)((cur).size()) - 1) {
        fr = cur[i + 1].first.first - cur[i].first.second;
      } else {
        fr = m - cur[i].first.second;
      }
      pair<int, int> rr = make_pair(0, 0);
      rr.first = cur[i].first.second + 1;
      if (i < (int)((cur).size()) - 1)
        rr.second = cur[i + 1].first.first;
      else
        rr.second = m;
      good.push_back(rr);
      cs.push_back(cur[i].second);
      while (sum(cs) > 3) {
        good.erase(good.begin());
        cs.erase(cs.begin());
      }
      for (int _n(((int)((good).size()) - 1) - 1), j(0); j <= _n; j++)
        ans += (long long)fr * (good[j].second - good[j].first + 1);
    }
  }
  return ans;
}
int main() {
  gets(mas[0]);
  sscanf(mas[0], "%d%d%d", &n, &m, &k);
  for (int _n((n)-1), i(0); i <= _n; i++) gets(mas[i]);
  k = 0;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    for (int _n((m)-1), j(0); j <= _n; j++) {
      if (mas[i][j] == '*' && mas[i + 1][j] != '*' && mas[i][j + 1] != '*') {
        int ci = i, cj = j;
        while (ci > 0 && mas[ci - 1][cj] == '*') --ci;
        while (cj > 0 && mas[ci][cj - 1] == '*') --cj;
        rect r;
        r.i1 = ci;
        r.i2 = i;
        r.j1 = cj;
        r.j2 = j;
        ob[k++] = r;
      }
    }
  }
  sort(ob, ob + k, cmp);
  c = 0;
  for (int _n((k)-1), i(0); i <= _n; i++) {
    beg[c++] = ob[i].i1;
    beg[c++] = ob[i].i2;
  }
  beg[c++] = 0;
  sort(beg, beg + c);
  c = unique(beg, beg + c) - beg;
  ans = 0;
  for (int _n((c)-1), ir(0); ir <= _n; ir++) {
    int r = beg[ir] + 1;
    long long tmp = 0;
    for (int _n((c)-1), il(0); il <= _n; il++) {
      int l = beg[il];
      if (l >= r) break;
      long long cur = calc(l, r);
      if (il)
        tmp += cur * (l - beg[il - 1]);
      else
        tmp += cur * (l + 1);
    }
    if (ir != c - 1)
      ans += tmp * (beg[ir + 1] - r + 1);
    else
      ans += tmp * (n - r + 1);
  }
  cout << ans << endl;
}