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 "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"

using namespace std;

const long long int MOD = 1000000007;

long long int N, M, K, H, W, L, R;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> H >> W;
	vector<string>s(H);
	for (auto &i : s)cin >> i;
	bitset<300>v[300];
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			v[i][j] = s[i][j] == '#';
		}
	}
	long long int ans = 0;
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			if (!v[i][j])continue;
			for (int k = i + 1, l = j + 1, cnt = 2; k < H&&l < W; k++, l++, cnt += 2) {
				if (!v[k][l])continue;
				for (int m = k - cnt-min(0,k-cnt), n = l-min(0,k-cnt); m <= i&&n < W; m++, n++) {
					if (v[m][n])ans++;
				}
				for (int m = k-min(0,l-cnt), n = l - cnt-min(0,l-cnt); m < H&&n <= j; m++, n++) {
					if (v[m][n])ans++;
				}
			}
			for (int k = i - 1, l = j + 1, cnt = 2; k >= 0 &&l < W; k--, l++, cnt += 2) {
				if (!v[k][l])continue;
				for (int m = k + cnt-1-max(0,k+cnt-(int)H), n = l+1+ max(0, k + cnt - (int)H); m > i&&n < W; m--, n++) {
					if (v[m][n])ans++;
				}
				for (int m = k - 1+ min(l - cnt + 1, 0), n = l - cnt + 1-min(l-cnt+1,0); m >= 0 && n < j; m--, n++) {
					if (v[m][n])ans++;
				}
			}
		}
	}
	cout << ans << endl;
	return 0;

}