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 <iostream>
#include <cstdio>
using namespace std;
int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}

char s[2003];
int pres[2003][2003], preh[2003][2003];
int q(int x, int y) {
    return preh[x][y] - preh[x][y - 1];
}

int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; ++i) {
        scanf("%s", s + 1);
        for (int j = 1; j <= m; ++j) {
            if (s[j] == '#') {
                pres[m + i - j][i + j]++;
                preh[m + i - j][i + j]++;
            }
        }
    }
    for (int j = 2; j <= m + n; ++j) {
        for (int i = 1; i <= m + n - 1; ++i) {
            pres[i][j] += pres[i - 1][j];
            preh[i][j] += preh[i][j - 1];
        }
    }
    long long res = 0;
    for (int k = 2; k <= m + n - 1; k ++) {
        for (int i = 1; i + k - 1 <= m + n - 1; ++i) {
            for (int j = 2; j + k - 1 <= m + n; ++j) {
                int sum = 0;
                if (q(i, j) && q(i + k - 1, j)) {
                    res += pres[i + k - 1][j + k - 1] - pres[i - 1][j + k - 1];
                    sum++;
                }
                if (q(i, j + k - 1) && q(i, j)) {
                    res += preh[i + k - 1][j + k - 1] - preh[i + k - 1][j - 1];
                    sum++;
                }
                if (q(i + k - 1, j) && q(i + k - 1, j + k - 1)) {
                    res += preh[i][j + k - 1] - preh[i][j - 1];
                    sum++;
                }
                if (q(i + k - 1, j + k - 1) && q(i, j + k - 1)) {
                    res += pres[i + k - 1][j] - pres[i - 1][j];
                    sum++;
                }
                if (sum == 2) res -= 1;
                if (sum == 4) res -= 4;
            }
        }
    }
    printf("%lld\n", res);
    return 0;
}