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
#include <stdio.h>
int main()
{
	char b[305][305];
	for (int i = 0; i < 305; i++) {
		for (int j = 0; j < 305; j++) {
			b[i][j] = '.';
		}
	}
	int h, w;
	scanf("%d %d", &h, &w);
	for (int i = 0; i < h; i++) {
		scanf("%s", b[i]);
	}
	int a[600][600];
	for (int i = 0; i < 600; i++) {
		for (int j = 0; j < 600; j++) {
			a[i][j] = 0;
		}
	}
	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			if (b[i][j] == '#') {
				a[i - j + 300][i + j + 1] = 1;
			}
		}
	}
	int r[600][600];
	for (int i = 0; i < 600; i++) {
		r[i][0] = 0;
		for (int j = 1; j < 600; j++) {
			if (a[i][j])r[i][j] = r[i][j - 1] + 1;
			else r[i][j] = r[i][j - 1];
		}
	}
	int y[600][600];
	for (int i = 0; i < 600; i++) {
		y[i][0] = 0;
		for (int j = 1; j < 600; j++) {
			if (a[j][i])y[i][j] = y[i][j - 1] + 1;
			else y[i][j] = y[i][j - 1];
		}
	}
	int u, v;
	long long ans = 0;
	for (int i = 0; i < 600; i++) {
		for (int j = 0; j < 600; j++) {
			for (int k = j + 1; k < 600; k++) {
				if (a[i][j] && a[i][k]) {
					u = i - (k - j);
					v = i + (k - j);
					if (u >= 0) {
						ans += (r[u][k-1] - r[u][j-1]);
					}
					if (v < 600) {
						ans += (r[v][k] - r[v][j]);
					}
				}
				if (a[j][i] && a[k][i]) {
					u = i - (k - j);
					v = i + (k - j);
					if (u >= 0) {
						ans += (y[u][k] - y[u][j]);
					}
					if (v < 600) {
						ans += (y[v][k - 1] - y[v][j - 1]);
					}
				}
			}
		}
	}
	printf("%lld\n", ans);
}