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<bits/stdc++.h>
using namespace std;
struct point{
	int x,y;
}p[90010];
int h,w,m=300,cnt,a[610][610],r[610][610],c[610][610];
long long ans;
int main()
{
	scanf("%d%d",&h,&w);
	for (int i=1;i<=h;i++)
	{
		char str[310];
		scanf("%s",str+1);
		for (int j=1;j<=w;j++)
		{
			if (str[j]=='#')
			{
				int x=i+j-1,y=i-j+m;
				p[cnt++]=(point){x,y};
				a[x][y]=1;
			}
		}
	}
	m*=2;
	for (int i=1;i<=m;i++)
	{
		for (int j=1;j<=m;j++)
		{
			r[i][j]=r[i][j-1]+a[i][j];
			c[i][j]=c[i-1][j]+a[i][j];
		}
	}
	for (int i=0;i<cnt;i++)
	{
		int x=p[i].x;
		int y=p[i].y;
		for (int d=2;d<=m;d+=2)
		{
			if (y+d<=m && a[x][y+d])
			{
				if (x+d<=m)
				{
					ans+=r[x+d][y+d]-r[x+d][y-1];
				}
				if (x-d>0)
				{
					ans+=r[x-d][y+d]-r[x-d][y-1];
				}
			}
			if (x+d<=m && a[x+d][y])
			{
				if (y+d<=m)
				{
					ans+=c[x+d-1][y+d]-c[x][y+d];
				}
				if (y-d>0)
				{
					ans+=c[x+d-1][y-d]-c[x][y-d];
				}
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}
