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 <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

char str[305][305];
int sum[2][305][605];

int main() {
  int n,m;
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++) scanf("%s",str[i]+1);
  ll ans=0;
  for(int i=1;i<=n;i++) {
  	for(int j=1;j<=m;j++) {
  		int d1=(i-1)+(j-1);
  		for(int k=1;k<=d1;k++) sum[0][j][k]=sum[0][j-1][k-1];
  		for(int k=1;k<i;k++)
  		  if (str[k][j-1]=='#') sum[0][j][i-k+1]++;
  		for(int k=1;i+k<=n&&j+k<=m&&k<=d1;k++)
  		  if (str[i+k][j]=='#'&&str[i][j+k]=='#') ans+=sum[0][j][k];
  		int d2=(n-i)+(j-1);
  		for(int k=1;k<=d2;k++) sum[1][j][k]=sum[1][j-1][k-1];
  		for(int k=i+1;k<=n;k++)
  		  if (str[k][j-1]=='#') sum[1][j][k-i+1]++;
  		for(int k=1;i-k>0&&j+k<=m&&k<=d2;k++)
  		  if (str[i-k][j]=='#'&&str[i][j+k]=='#') ans+=sum[1][j][k];
	  }
	for(int j=m;j>0;j--) {
		int d1=(i-1)+(m-j);
		for(int k=1;k<=d1;k++) sum[0][j][k]=sum[0][j+1][k-1];
		for(int k=1;k<i;k++)
		  if (str[k][j+1]=='#') sum[0][j][i-k+1]++;
		for(int k=1;i+k<=n&&j-k>0&&k<=d1;k++) 
		  if (str[i+k][j]=='#'&&str[i][j-k]=='#') ans+=sum[0][j][k];
		int d2=(n-i)+(m-j);
		for(int k=1;k<=d2;k++) sum[1][j][k]=sum[1][j+1][k-1];
		for(int k=i+1;k<=n;k++)
		  if (str[k][j+1]=='#') sum[1][j][k-i+1]++;
		for(int k=1;i-k>0&&j-k>0&&k<=d2;k++)
		  if (str[i-k][j]=='#'&&str[i][j-k]=='#') ans+=sum[1][j][k];
	}
  }
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
      for(int k=1;k<=min(n,m);k++) {
      	if (i-k<=0&&i+k>n&&j-k<=0&&j+k>m) break;
      	if (i-k>0&&i+k<=n&&j-k>0&&str[i-k][j]=='#'&&str[i+k][j]=='#'&&str[i][j-k]=='#') ans++;
      	if (i-k>0&&i+k<=n&&j+k<=m&&str[i-k][j]=='#'&&str[i+k][j]=='#'&&str[i][j+k]=='#') ans++;
      	if (i-k>0&&j-k>0&&j+k<=m&&str[i-k][j]=='#'&&str[i][j-k]=='#'&&str[i][j+k]=='#') ans++;
      	if (i+k<=n&&j-k>0&&j+k<=m&&str[i+k][j]=='#'&&str[i][j-k]=='#'&&str[i][j+k]=='#') ans++;
	  }
  printf("%lld\n",ans);
  return 0;
}