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
import itertools
import sys
H, W = [int(i) for i in input().split(' ')]

cross = [[0] * (H + W -1) for _ in range(H + W - 1)]
for i, line in enumerate(sys.stdin):
    for j, c in enumerate(line.strip()):
        if c == "#":
            cross[i+j][i-j+W-1] = 1

across = [[0] + list(itertools.accumulate(xs)) for xs in cross]
r_cross = [[xs[i] for xs in cross] for i in range(H + W - 1)]
across2 = [[0] + list(itertools.accumulate(xs)) for xs in r_cross]

r = 0
for i in range(H+W-1):
    for j in range(H+W-1):
        if cross[i][j] != 1:
            continue
        for k in range(j+1, H+W-1):
            if cross[i][k] == 1:
                if i - (k-j) >= 0:
                    r += across[i - (k - j)][k+1] - across[i - (k - j)][j]
                if i + (k-j) < H + W - 1:
                    r += across[i + (k - j)][k+1] - across[i + (k - j)][j]

for i in range(H+W-1):
    for j in range(H+W-1):
        if cross[j][i] != 1:
            continue
        for k in range(j+1, H+W-1):
            if cross[k][i] == 1:
                if i - (k-j) >= 0:
                    r += across2[i - (k - j)][k] - across2[i - (k - j)][j+1]
                if i + (k-j) < H + W - 1:
                    r += across2[i + (k - j)][k] - across2[i + (k - j)][j+1]

print(r)
