#include <bits/stdc++.h>
using namespace std;
int N, M, F[2][305][305];
char A[305][305];
long long Ans;
inline int LU(int x, int y, int k) {
  int cnt = 0;
  for (int i = 0; i < k; ++i) {
    if (x - i < 1 || y - i < 1) break;
    cnt += A[x - i][y - i] == '#';
  }
  return cnt;
  if (k < 1) return 0;
  k += min(0, min(x - k, y - k));
  return F[0][x][y] - F[0][x - k][y - k];
}
inline int RU(int x, int y, int k) {
  int cnt = 0;
  for (int i = 0; i < k; ++i) {
    if (x - i < 1 || y + i > M) break;
    cnt += A[x - i][y + i] == '#';
  }
  return cnt;
  if (k < 1) return 0;
  k += max(0, min(x - k, y + k - M - 1));
  return F[1][x][y] - F[1][x - k][y + k];
}
inline int Calc(int x_1, int y_1, int x_2, int y_2) {
  if (x_2 < x_1) swap(x_1, x_2), swap(y_1, y_2);
  if (y_1 <= y_2) {
    int z = max(x_2 - N, y_2 - M);
    if (z > 0) x_2 -= z, y_2 -= z;
    if (x_2 < 1 || y_2 < 1) return 0;
    return LU(x_2, y_2, x_2 - x_1 + 1);
  }
  if (y_1 >= y_2) {
    int z = max(x_2 - N, 1 - y_2);
    if (z > 0) x_2 -= z, y_2 += z;
    if (x_2 < 1 || y_2 > M) return 0;
    return RU(x_2, y_2, x_2 - x_1 + 1);
  }
}
int main() {
  scanf("%d%d", &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)
      F[0][i][j] = F[0][i - 1][j - 1] + (A[i][j] == '#'),
      F[1][i][j] = F[1][i - 1][j + 1] + (A[i][j] == '#');
  int KS = min(N, M);
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= M; ++j)
      for (int k = KS; k; --k) {
        if (i > k && j > k && A[i - k][j] == '#' && A[i][j - k] == '#')
          Ans += Calc(i - k, j - 2 * k, i - 2 * k + 1, j - k - 1);
        if (i + k <= N && j + k <= M && A[i + k][j] == '#' &&
            A[i][j + k] == '#')
          Ans += Calc(i + k, j + 2 * k, i + 2 * k - 1, j + k + 1);
        if (i > k && j + k <= M && A[i - k][j] == '#' && A[i][j + k] == '#')
          Ans += Calc(i - 2 * k, j + k, i - k - 1, j + 2 * k - 1);
        if (i + k <= N && j > k && A[i + k][j] == '#' && A[i][j - k] == '#')
          Ans += Calc(i + 2 * k, j - k, i + k + 1, j - 2 * k + 1);
      }
  printf("%d\n", Ans);
  return 0;
}
