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
H,W = map(int,input().split())
src = [input() for i in range(H)]

def solve():
    cums = [[0]*(H+W) for y in range(H+W-1)]
    coins = [[] for y in range(H+W-1)]
    for i in range(H):
        for j in range(W):
            if src[i][j] == '.': continue
            x,y = j-i+H, j+i
            cums[y][x] += 1
            coins[y].append(x)
    for i in range(H+W-1):
        for j in range(H+W-1):
            cums[i][j+1] += cums[i][j]
    cnt = 0
    for y1 in range(H+W-1):
        if len(coins[y1]) < 2: continue
        for i1 in range(len(coins[y1])-1):
            for i2 in range(i1+1,len(coins[y1])):
                x1,x2 = coins[y1][-1-i1], coins[y1][-1-i2]
                y2 = y1 - (x2 - x1)
                if y2 < 0: continue
                cnt += cums[y2][x2] - cums[y2][x1]
    return cnt

def rotate():
    global H,W,src
    H,W = W,H
    src2 = []
    for col in reversed(list(zip(*src))):
        src2.append(''.join(list(col)))
    src = src2

ans = 0
for i in range(4):
    rotate()
    ans += solve()
print(ans)