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<cstdio>
#include<algorithm>
#include<cstring>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
using namespace std;

const int N=610;
struct point{ int x,y; }p[90010];
int h,w,m=300,cnt,a[N][N],r[N][N],c[N][N];
long long ans;

int main(){
	scanf("%d%d",&h,&w);
	rep(i,1,h){
		char str[310]; scanf("%s",str+1);
		rep(j,1,w){
			if (str[j]=='#'){
				int x=i+j-1,y=i-j+m;
				p[cnt++]=(point){x,y}; a[x][y]=1;
			}
		}
	}
	m*=2;
	rep(i,1,m) rep(j,1,m) 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,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];
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}