Problem: There are some coins in the xy-plane. The positions of the coins are represented by a grid of characters with H rows and W columns. If the character at the i-th row and j-th column, s_{ij}, is `#`, there is one coin at point (i,j); if that character is `.`, there is no coin at point (i,j). There are no other coins in the xy-plane.

There is no coin at point (x,y) where 1\leq i\leq H,1\leq j\leq W does not hold. There is also no coin at point (x,y) where x or y (or both) is not an integer. Additionally, two or more coins never exist at the same point.

Find the number of triples of different coins that satisfy the following condition:

* Choosing any two of the three coins would result in the same Manhattan distance between the points where they exist.



Here, the Manhattan distance between points (x,y) and (x',y') is |x-x'|+|y-y'|. Two triples are considered the same if the only difference between them is the order of the coins.

Constraints

* 1 \leq H,W \leq 300
* s_{ij} is `#` or `.`.

Input

Input is given from Standard Input in the following format:


H W
s_{11}...s_{1W}
:
s_{H1}...s_{HW}


Output

Print the number of triples that satisfy the condition.

Examples

Input

5 4
#.##
.##.
#...
..##
...#


Output

3


Input

5 4
.##
.##.
...
..##
...#


Output

3


Input

13 27
......#.........#.......#..
...#.....###..
..............#####...##...
...#######......#...#######
...#.....#.....###...#...#.
...#######....#.#.#.#.###.#
..............#.#.#...#.#..
.#.#.#...###..
...........#...#...#######
..#######..#...#...#.....#
..#.....#..#...#...#.###.#
..#######..#...#...#.#.#.#
..........##...#...#.#####


Output

870

Solution (in CPP):

Solution:
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <utility>
#include <cstdio>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#define my_abs(x) ((x) < 0 ? -(x) : (x))
#define mp std::make_pair
#define pb push_back
typedef long long ll;
int row[605][605], col[605][605], app[605];
bool arr[605][605];
char str[605];
int main()
{
	// freopen("TK1-E.in", "r", stdin);
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
	{
		scanf("%s", str);
		for (int j = 0; j < m; j++)
		{
			if (str[j] == '#')
				arr[i + j][i - j + m] = true;
		}
	}
	n += m;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			row[i][j] = (j ? row[i][j - 1] : 0) + arr[i][j];
			col[i][j] = (i ? col[i - 1][j] : 0) + arr[i][j];
		}
	}
	ll ans = 0;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (!arr[i][j])
				continue;
			for (int k = j + 1; k < n; k++)
			{
				if (!arr[i][k])
					continue;
				int d = k - j;
				if (i + d < n)
					ans += row[i + d][k - 1] - row[i + d][j];
				if (i - d >= 0)
					ans += row[i - d][k - 1] - row[i - d][j];
			}
		}
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (!arr[i][j])
				continue;
			for (int k = i + 1; k < n; k++)
			{
				if (!arr[k][j])
					continue;
				int d = k - i;
				if (j + d < n)
					ans += col[k - 1][j + d] - col[i][j + d];
				if (j - d >= 0)
					ans += col[k - 1][j - d] - col[i][j - d];
			}
		}
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (!arr[i][j])
				continue;
			memset(app, 0, sizeof(app));
			for (int k = 0; k < n; k++)
			{
				if (j == k || !arr[i][k])
					continue;
				app[my_abs(k - j)]++;
			}
			for (int k = 0; k < n; k++)
			{
				if (i == k || !arr[k][j])
					continue;
				ans += app[my_abs(k - i)];
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}