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 <bits/stdc++.h>

using namespace std;

int n, m;
long long y[605], z[605], ans;
char x[305][305];
int ta;

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
		scanf("%s", x[i]);
	for (int i = 0; i < n; i++) {
		memset(y, 0, sizeof(y));
		memset(z, 0, sizeof(z));
		for (int j = 0; j < i; j++)
			for (int k = 0; k < m; k++)
				y[abs(j - i) + k] += x[j][k] == '#';
		for (int j = i + 1; j < n; j++)
			for (int k = 0; k < m; k++)
				z[abs(j - i) + k] += x[j][k] == '#';
		for (int j = 0; j < m; j++) {
			for (int k = 0; k < i; k++)
				y[abs(k - i) + j] -= x[k][j] == '#';
			for (int k = i + 1; k < n; k++)
				z[abs(k - i) + j] -= x[k][j] == '#';
			for (int k = 1; j - k >= 0 && i + k < n; k++)
				if (x[i][j - k] == '#' && x[i + k][j] == '#')
				  ans += y[k + j];
			for (int k = 1; j - k >= 0 && i - k >= 0; k++)
				if (x[i][j - k] == '#' && x[i - k][j] == '#')
				  ans += z[k + j];
			for (int k = 1; k < 305; k++) {
				ta = 0;
				if (i - k >= 0)
					ta += x[i - k][j] == '#';
				if (j - k >= 0)
					ta += x[i][j - k] == '#';
				if (i + k < n)
					ta += x[i + k][j] == '#';
				if (j + k < m)
					ta += x[i][j + k] == '#';
				if (ta == 3)
					ans++;
				else if (ta == 4)
					ans += 4;
			}
		}
	}
	for (int i = 0; i < n; i++)
		reverse(x[i], x[i] + m);
	for (int i = 0; i < n; i++) {
		memset(y, 0, sizeof(y));
		memset(z, 0, sizeof(z));
		for (int j = 0; j < i; j++)
			for (int k = 0; k < m; k++)
				y[abs(j - i) + k] += x[j][k] == '#';
		for (int j = i + 1; j < n; j++)
			for (int k = 0; k < m; k++)
				z[abs(j - i) + k] += x[j][k] == '#';
		for (int j = 0; j < m; j++) {
			for (int k = 0; k < i; k++)
				y[abs(k - i) + j] -= x[k][j] == '#';
			for (int k = i + 1; k < n; k++)
				z[abs(k - i) + j] -= x[k][j] == '#';
			for (int k = 1; j - k >= 0 && i + k < n; k++)
				if (x[i][j - k] == '#' && x[i + k][j] == '#')
				  ans += y[k + j];
			for (int k = 1; j - k >= 0 && i - k >= 0; k++)
				if (x[i][j - k] == '#' && x[i - k][j] == '#')
				  ans += z[k + j];
		}
	}
	printf("%lld\n", ans);
	return 0;
}