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;

typedef long long ll;
const int maxn=300+10;
int n,m;
char s[maxn][maxn];
ll ans;

inline void work(){
	static int a[maxn<<1];
	for(int i=1,x,y,z;i<=n;++i,memset(a,0,sizeof(a)))
		for(int j=1;j<=m;++j){
			x=j-1,y=j+1,z=i-1;
			while(x&&y<=m&&z){
				if(s[i][x]=='#'&&s[i][y]=='#'&&s[z][j]=='#')
					++ans;
				--x;++y;--z;
			}
			x=j+1,y=i+1;
			while(x<=m&&y<=n){
				if(s[i][x]=='#'&&s[y][j]=='#'&&i+j-x+j>=0&&i+j-x+j<maxn<<1)
					ans+=a[i+j-x+j];
				++x;++y;
			}
			for(int k=1;k<i;++k)
				if(s[k][j]=='#')
					++a[k+j];
		}
}
inline void rotate(){
	static char t[maxn][maxn];
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
			t[m-j+1][i]=s[i][j];
	swap(n,m);
	memcpy(s,t,sizeof(s));
}

int main(){
	cin>>n>>m;
	for(int i=1;i<=n;++i)
		cin>>s[i]+1;
	for(int i=1;i<=4;++i,rotate())
		work();
	cout<<ans<<endl;
	return 0;
}