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;

int H, W;
int fid[610][610], cmp[610][610];
long long ans;

int CMP(int i, int j)
{
    if(j < 0)return 0;
    return cmp[i][j];
}

void unit()
{
    memset(cmp, 0, sizeof(cmp));
    for(int i = 0; i < 610; i++)for(int j = 0; j < 610; j++)
    {
        if(j == 0)cmp[i][j] = fid[i][j];
        else cmp[i][j] = fid[i][j] + cmp[i][j - 1];
    }

    for(int i = 0; i < 610; i++)
    {
        for(int j = 0; j < 610; j++)
        {
            for(int k = j + 1; k < 610; k++)
            {
                if(fid[i][j] && fid[i][k])
                {
                    int d = k - j;
                    if(i + d < 610)ans += 2 * (CMP(i + d, k - 1) - CMP(i + d, j));
                    if(i - d >= 0)ans += 2 * (CMP(i - d, k - 1) - CMP(i - d, j));
                    if(i + d < 610 && fid[i + d][j])ans++;
                    if(i - d >= 0 && fid[i - d][j])ans++;
                    if(i + d < 610 && fid[i + d][k])ans++;
                    if(i - d >= 0 && fid[i - d][k])ans++;
                }
            }
        }
    }
    return;
}

int main()
{
    cin >> H >> W;
    for(int i = 0; i < H; i++)for(int j = 0; j < W; j++)
    {
        int x = i + j, y = i - j + 300;
        char c;
        cin >> c;
        fid[x][y] = (c == '#');
    }

    unit();

    int cpy[610][610] = {0};
    for(int i = 0; i < 610; i++)
    {
        for(int j = 0; j < 610; j++)
        {
            cpy[j][i] = fid[i][j];
        }
    }
    for(int i = 0; i < 610; i++)for(int j = 0; j < 610; j++)fid[i][j] = cpy[i][j];

    unit();

    cout << ans / 2 << endl;

    return 0;
}