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;

#define FOR(a, b, c) for(int a = b; a <= c; ++a)
const int N = 305;

int n, m, ans;
bitset<305> a[N];

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> n >> m;
    for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) {
        char x; cin >> x;
        if(x == '#') a[i].set(j);
    }
    FOR(i, 1, n) FOR(j, i, n) FOR(t, j, n)  if(i != t)  {
        bitset<305> a1 = a[i], a2 = a[j], a3 = a[t];
        int d1 = j - i, d2 = t - j;
        int s1 = d1, s2 = 2*d1 - d2;
        if(d1 <= d2) s1 = 2*d2 - d1, s2 = d2;
        a1 <<= s1; a3 <<= s2;
        a2 &= a1; a2 &= a3;
        ans += a2.count();

        if(d1 == 0 || d2 == 0) continue;
        a1 = a[i], a2 = a[j], a3 = a[t];
        a1 >>= s1; a3 >>= s2;
        a2 &= a1; a2 &= a3;
        ans += a2.count();
    }
    cout << ans;
}