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 N = 305;
int n, m, cnt[3 * N][3 * N], ans;
char c[3 * N][3 * N], cc[3 * N][3 * N];

bool valid (int x, int y) { return x > 0 && y > 0 && x <= n && y <= m; }

void rotation () {
    for (int i = 0; i < 3 * N; ++i) {
        for (int j = 0; j < 3 * N; ++j) {
            cc[i][j] = '.';
        }
    }
    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= n; ++j) {
            cc[i + m][j + n] = c[j + n][m + 1 - i + m];
        }
    }
    swap(n, m);
    for (int i = 0; i < 3 * N; ++i) {
        for (int j = 0; j < 3 * N; ++j) {
            c[i][j] = '.';
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            c[i + n][j + m] = cc[i + n][j + m];
        }
    }
}

void solve () {
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= 3 * n; ++i) {
        for (int j = 1; j <= 3 * m; ++j) {
            cnt[i][j] = cnt[i - 1][j + 1] + (c[i][j] == '#');
        }
    }
    for (int x = 1; x <= n; ++x) {
        for (int y = 1; y <= m; ++y) {
            for (int d = 1; d <= n; ++d) {
                int X = x - d, Y = y + d;
                if (!valid(X, Y) || c[x + n][y + m] == '.' || c[X + n][Y + m] == '.') continue ;
                ans += cnt[x - d + n][y - d + m] - cnt[x - 2 * d + n][y + m];
            }
        }
    }
}

int main () {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            cin >> c[i + n][j + m];
        }
    }

    for (int _ = 0; _ < 4; ++_) {
        solve();
        rotation();
    }

    cout << ans;

    return 0;
}