#include <bits/stdc++.h>
using namespace std;
long long a[1000010], b[1000010], e[1001][1001], o[300010], prime[300010], n1,
    n2, t, l, r, k, c, x, y, z, j, c1, c2, n, d, m, sum, fact[300010];
stack<long long> s;
deque<long long> q;
string st, st1, st2;
set<long long> se, se1;
vector<long long> adj[100010], visited, dis;
map<long long, long long> mp;
vector<pair<long long, long long>> we[100010];
pair<long long, pair<long long, long long>> p[300010], p1[300010];
char ch[1001][1001];
const long long mod = 1e9 + 7;
multimap<long long, long long> mm;
multiset<long long> ms1, ms2;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      cin >> ch[i][j];
      if (ch[i][j] == '*') a[i]++, b[j]++;
    }
  for (long long i = 0; i < n; i++)
    for (j = 0; j < m; j++)
      if (ch[i][j] == '*') sum += (a[i] - 1) * (b[j] - 1);
  cout << sum;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long a[1000010], b[1000010], e[1001][1001], o[300010], prime[300010], n1,
    n2, t, l, r, k, c, x, y, z, j, c1, c2, n, d, m, sum, fact[300010];
stack<long long> s;
deque<long long> q;
string st, st1, st2;
set<long long> se, se1;
vector<long long> adj[100010], visited, dis;
map<long long, long long> mp;
vector<pair<long long, long long>> we[100010];
pair<long long, pair<long long, long long>> p[300010], p1[300010];
char ch[1001][1001];
const long long mod = 1e9 + 7;
multimap<long long, long long> mm;
multiset<long long> ms1, ms2;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      cin >> ch[i][j];
      if (ch[i][j] == '*') a[i]++, b[j]++;
    }
  for (long long i = 0; i < n; i++)
    for (j = 0; j < m; j++)
      if (ch[i][j] == '*') sum += (a[i] - 1) * (b[j] - 1);
  cout << sum;
  return 0;
}

```