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 <iostream>
using namespace std;
typedef long long ll;

const int M=620,M2=310;
int h,w;
int a[M][M],b[M][M];

ll f(int B){
	ll res=0;
	for(int i=0;i<M;i++) for(int j=1;j<M;j++) b[i][j]=a[i][j]+b[i][j-1];
	for(int i=0;i<M;i++) for(int j=0;j<M;j++) if(a[i][j]){
		for(int k=j+1;k<M;k++) if(a[i][k]){
			int d=k-j;
			if(i+d<M) res+=b[i+d][k-B]-b[i+d][j-1+B];
			if(d<=i) res+=b[i-d][k-B]-b[i-d][j-1+B];
		}
	}
	for(int i=0;i<M;i++) for(int j=0;j<M;j++) b[M-j-1][i]=a[i][j];
	for(int i=0;i<M;i++) for(int j=0;j<M;j++) a[i][j]=b[i][j],b[i][j]=0;
	return res;
}

int main(){
	cin>>h>>w;
	for(int i=0;i<h;i++) for(int j=0;j<w;j++){
		char c;
		cin>>c;
		if(c=='#') a[i+j+1][i-j+M2+1]++;
	}
	cout<<f(0)+f(1)<<endl;
}