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;
const int M = 1000000007;

int main() {
    int H, W;
    cin >> H >> W;
    vector<string> S(H);
    int h = H + W - 1;
    int w = H + W;
    vector<vector<int>> s(h, vector<int>(w, 0));
    for (int i = 0; i < H; ++i) {
        cin >> S[i];
        for (int j = 0; j < W; ++j) {
            if (S[i][j] == '#') {
                s[i + j][i - j + W] = 1;
            }
        }
    }
    vector<vector<int>> imsw(h, vector<int>(w + 1, 0));
    vector<vector<int>> imsh(h + 1, vector<int>(w, 0));
    for (int i = 0; i < h; ++i) {
        for (int j = 0; j < w; ++j) {
            imsw[i][j + 1] = imsw[i][j] + s[i][j];
        }
    }
    for (int j = 0; j < w; ++j) {
        for (int i = 0; i < h; ++i) {
            imsh[i + 1][j] = imsh[i][j] + s[i][j];
        }
    }
    long long ans = 0;
    for (int i = 0; i < h; ++i) {
        for (int j = 0; j < w; ++j) {
            if (!s[i][j]) continue;
            for (int ti = i + 1; ti < h; ++ti) {
                if (s[ti][j]) {
                    int d = ti - i;
                    if (j >= d)
                        ans += imsh[ti + 1][j - d] - imsh[i][j - d];
                    if (j + d < w)
                        ans += imsh[ti + 1][j + d] - imsh[i][j + d];
                }
            }
            for (int tj = j + 1; tj < w; ++tj) {
                if (s[i][tj]) {
                    int d = tj - j;
                    if (i >= d)
                        ans += imsw[i - d][tj + 1] - imsw[i - d][j];
                    if (i + d < h)
                        ans += imsw[i + d][tj + 1] - imsw[i + d][j];
                }
            }
        }
    }
    for (int i = 0; i < h; ++i) {
        for (int j = 0; j < w; ++j) {
            if (!s[i][j]) continue;
            for (int k = 1; k < w; ++k) {
                int a = 0, b = 0;
                if (i >= k && s[i - k][j])
                    ++a;
                if (i + k < h && s[i + k][j])
                    ++a;
                if (j >= k && s[i][j - k])
                    ++b;
                if (j + k < w && s[i][j + k])
                    ++b;

                ans -= a * b;
            }
        }
    }
    cout << ans << "\n";
    return 0;
}