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>
#include <sys/types.h>
#include <unistd.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

using ll=long long;
using R=long double;
const R EPS=1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r){return(r > EPS)-(r < -EPS);}
inline R sq(R x){return sqrt(max(x,0.0L));}

const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};

const pid_t pid = getpid();
// Problem Specific Parameter:

const int offset = 300;
char board[600][600];
int hsum[610][610];
int vsum[610][610];
using B = bitset<610>;

int main(void){
	int h,w;
	cin >> h >> w;

	const int n = 601;
	rep(i,n)rep(j,n) board[i][j] = '.';

	rep(y,h)rep(x,w){
		char tmp;
		cin >> tmp;
		const int ny = x + y + 1;
		const int nx = x - y + offset + 1;
		board[ny][nx] = tmp;
	}

	rep(y,n){
		rep(x,1,n){
			hsum[y][x] = (board[y][x] == '#');
			hsum[y][x] += hsum[y][x-1]; 
		}
	}

	rep(x,n){
		rep(y,1,n){
			vsum[y][x] = (board[y][x] == '#');
			vsum[y][x] += vsum[y-1][x]; 
		}
	}

	ll ans = 0LL;
	rep(y,n){
		rep(x2,n)rep(x,x2){
			if(board[y][x] == '.') continue;
			if(board[y][x2] == '.') continue;
			const int d = abs(x2 - x);
			
			if(y - d > 0){
				ans += 1LL * (hsum[y - d][x2] - hsum[y - d][x - 1]);
			}

			if(y + d < n){
				ans += 1LL * (hsum[y + d][x2] - hsum[y + d][x - 1]);
			}
		} 
	}

	rep(x,n){
		rep(y2,n)rep(y,y2){
			if(board[y][x] == '.') continue;
			if(board[y2][x] == '.') continue;
			const int d = abs(y2 - y);
			
			if(x - d > 0){
				ans += 1LL * (vsum[y2 - 1][x - d] - vsum[y][x - d]);
			}

			if(x + d < n){
				ans += 1LL * (vsum[y2 - 1][x + d] - vsum[y][x + d]);
			}
		} 
	}


	cout << ans << endl;

	return 0;
}