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;
#define rep(i,n) for(int i=0; i<(n); i++)
int h, w, b[605][605], c[605][605], d[605][605];
long long ans;
char s[305];
int main(){
	scanf("%d %d", &h, &w);
	rep(i,h){
		scanf("\n%s", s);
		rep(j,w) if(s[j] == '#') b[i+j][i-j+w] = 1;
	}
	rep(i,h+w) rep(j,h+w){
		c[i][j+1] = c[i][j] + b[i][j];
		d[i+1][j] = d[i][j] + b[i][j];
	}
	rep(i,h+w) rep(j,h+w) if(b[i][j]) rep(k,h+w){
		if(j+k < h+w && b[i][j+k]) ans += (i >= k ? c[i-k][j+k] - c[i-k][j] : 0) + (i+k < h+w ? c[i+k][j+k] - c[i+k][j] : 0);
		if(i+k < h+w && b[i+k][j]) ans += (j >= k ? d[i+k+1][j-k] - d[i][j-k] : 0) + (j+k < h+w ? d[i+k][j+k] - d[i+1][j+k] : 0);
	}
	printf("%lld\n", ans);
}