#include <bits/stdc++.h>
using namespace std;
template <class T, class L>
bool smax(T &x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
bool smin(T &x, L y) {
  return y < x ? (x = y, 1) : 0;
}
const int maxn = 1e3 + 17;
int n, m, row[maxn], col[maxn];
bool c[maxn][maxn];
int main() {
  scanf("%d %d\n", &n, &m);
  for (int i = 0; i < n; i++, getchar())
    for (int j = 0; j < m; j++)
      if (c[i][j] = getchar() == '*') row[i]++, col[j]++;
  long long ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) ans += c[i][j] * (row[i] - 1) * (col[j] - 1);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

You are given a n × m field consisting only of periods ('.') and asterisks ('*'). Your task is to count all right triangles with two sides parallel to the square sides, whose vertices are in the centers of '*'-cells. A right triangle is a triangle in which one angle is a right angle (that is, a 90 degree angle).

Input

The first line contains two positive integer numbers n and m (1 ≤ n, m ≤ 1000). The following n lines consist of m characters each, describing the field. Only '.' and '*' are allowed.

Output

Output a single number — total number of square triangles in the field. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

2 2
**
*.


Output

1


Input

3 4
*..*
.**.
*.**


Output

9

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T, class L>
bool smax(T &x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
bool smin(T &x, L y) {
  return y < x ? (x = y, 1) : 0;
}
const int maxn = 1e3 + 17;
int n, m, row[maxn], col[maxn];
bool c[maxn][maxn];
int main() {
  scanf("%d %d\n", &n, &m);
  for (int i = 0; i < n; i++, getchar())
    for (int j = 0; j < m; j++)
      if (c[i][j] = getchar() == '*') row[i]++, col[j]++;
  long long ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) ans += c[i][j] * (row[i] - 1) * (col[j] - 1);
  printf("%lld\n", ans);
  return 0;
}

```