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;
using ll = long long;
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...) 0
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int h, w;
    cin >> h >> w;
    vector<string> a(h);
    rep(i, 0, h) cin >> a[i];
    ll ans = 0;
    vector<vector<int>> b(h + w, vector<int>(h + w));
    rep(i, 0, h) rep(j, 0, w) {
        if (a[i][j] == '#') b[i + 1 + j][h - i + j] = 1;
    }
    vector<vector<int>> c = b;
    rep(i, 1, h + w) rep(j, 1, h + w) {
        c[i][j] += c[i][j - 1];
    }
    rep(i, 1, h + w) {
        rep(j, 1, h + w) {
            if (b[i][j] != 1) continue;
            rep(k, j + 1, h + w) {
                if (b[i][k] != 1) continue;
                int dif = k - j;
                if (i + dif < h + w) ans += c[i + dif][k] - c[i + dif][j - 1];
                if (i - dif >= 0) ans += c[i - dif][k] - c[i - dif][j - 1];
                // debug(i, j, k, ans);
            }
        }
    }
    c = b;
    rep(i, 1, h + w) rep(j, 1, h + w) {
        c[i][j] += c[i - 1][j];
    }
    rep(j, 1, h + w) {
        rep(i, 1, h + w) {
            if (b[i][j] != 1) continue;
            rep(l, i + 1, h + w) {
                if (b[l][j] != 1) continue;
                int dif = l - i;
                if (j + dif < h + w) ans += c[l - 1][j + dif] - c[i][j + dif];
                if (j - dif >= 0) ans += c[l - 1][j - dif] - c[i][j - dif];
                // debug(i, j, l, ans);
            }
        }
    }
    cout << ans << endl;
    return 0;
}