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;

const int N = 606; 
int a[N][N], R[N][N], C[N][N];

void solve() {
    int h, w;
    cin >> h >> w;
    for (int i = 0; i < h; i++) {
        string s; cin >> s;
        for (int j = 0; j < w; j++) {
            if (s[j]=='#') a[i+j][i-j+w] = 1;
        }
    }
    int n = h+w;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            R[i][j+1] = R[i][j] + a[i][j];
            C[i+1][j] = C[i][j] + a[i][j];
        }
    }

    int res = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (a[i][j]) {
                for (int k = 0; k < n; k++) {
                    if (j+k < n && a[i][j+k]) {
                        res += (i-k<0?0: R[i-k][j+k+1] - R[i-k][j+1]);
                        res += (i+k>=n?0:R[i+k][j+k] - R[i+k][j]);
                    }
                    if (i+k < n && a[i+k][j]) {
                        res += (j-k<0?0: C[i+k][j-k] - C[i][j-k]);
                        res += (j+k>=n?0:C[i+k+1][j+k] - C[i+1][j+k]);
                    }
                }
            }
        }
    }
    cout << res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}
