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 <cstdio>
#include <cstring>

#include <algorithm>

using namespace std;

#define NMAX 600
#define BMAX (NMAX * 3 + 6)

typedef long long i64;

static int n, m;
static char map[NMAX + 10][NMAX + 10];
static int pre[3 * NMAX + 10][3 * NMAX + 10];

#define S(i, j) pre[NMAX + (i)][NMAX + (j)]


void show() {
    for (int i = 1; i <= n; i++)
        printf("%s\n", map[i] + 1);
    puts("");
}

void rotate() {
    static char buf[NMAX + 10][NMAX + 10];
    memcpy(buf, map, sizeof(buf));
    memset(map, 0, sizeof(map));
    for (int i = 1; i <= m; i++) for (int j = 1; j <= n; j++)
        map[i][j] = buf[n - j + 1][i];
    swap(n, m);
    // show();
}

i64 eval() {
    memset(pre, 0, sizeof(pre));
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++)
        S(i, j) = map[i][j] == '#';
    for (int i = 0; i <= BMAX; i++) for (int j = 1; i >= j; j++)
        pre[i - j][j] += pre[i - j + 1][j - 1];

    i64 r = 0;
    for (int i = 1; i <= n + m; i++)
        for (int j = 0; i - j >= 1; j++) if (map[i - j][j + 1] == '#')
        for (int k = j + 1; i - k >= 1; k++) if (map[i - k][k + 1] == '#') {
            int d = (k - j) << 1;
            r += S(i - j - d, j + 1) - S(i - k, k - d + 1);
        }
    return r;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%s", map[i] + 1);

    i64 ans = eval();
    rotate();
    ans += eval();
    rotate();
    ans += eval();
    rotate();
    ans += eval();
    printf("%lld\n", ans);

    return 0;
}