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 memo[605][605]={},sh[605][605]={},sw[605][605]={};
string s[305];

int main(int argc, char const *argv[])
{
	int h,w;
	cin>>h>>w;
	for (int i = 0; i < h; ++i)
	{
		cin>>s[i];
	}
	for (int i = 0; i < h; ++i)
	{
		for (int j = 0; j < w; ++j)
		{
			if (s[i][j]=='#')
			{
				memo[i+j+1][i-j+300]=1;
			}
		}
	}
	for (int i = 1; i < 605; ++i)
	{
		for (int j = 1; j < 605; ++j)
		{
			sh[i][j]=memo[i][j]+sh[i-1][j];
			sw[i][j]=memo[i][j]+sw[i][j-1];
		}
	}
	long long ans=0LL;
	for (int i = 1; i < 600; ++i)
	{
		for (int j = 1; j < 600; ++j)
		{
			for (int k = j+1; k < 600; ++k)
			{
				int d=k-j;
				if (memo[i][j]&&memo[i][k])
				{
					if (i-d>=0)
					{
						ans+=sw[i-d][k]-sw[i-d][j-1];
						ans-=memo[i-d][j]+memo[i-d][k];
					}
					if (i+d<=600)
					{
						ans+=sw[i+d][k]-sw[i+d][j-1];
						ans-=memo[i+d][j]+memo[i+d][k];
					}
				}
				if (memo[j][i]&&memo[k][i])
				{
					if (i-d>=0)
					{
						ans+=sh[k][i-d]-sh[j-1][i-d];
					}
					if (i+d<=600)
					{
						ans+=sh[k][i+d]-sh[j-1][i+d];
					}
				}
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}