Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

5 4
#.##
.##.
#...
..##
...#

Sample Output 1

3

((1,1),(1,3),(2,2)),((1,1),(2,2),(3,1)) and ((1,3),(3,1),(4,4)) satisfy the condition.

Sample Input 2

13 27
......#.........#.......#..
#############...#.....###..
..............#####...##...
...#######......#...#######
...#.....#.....###...#...#.
...#######....#.#.#.#.###.#
..............#.#.#...#.#..
#############.#.#.#...###..
#...........#...#...#######
#..#######..#...#...#.....#
#..#.....#..#...#...#.###.#
#..#######..#...#...#.#.#.#
#..........##...#...#.#####

Sample Output 2

870

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned int uint;
typedef unsigned char uchar;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

#define REP(i,x) for(int i=0;i<(int)(x);i++)
#define REPS(i,x) for(int i=1;i<=(int)(x);i++)
#define RREP(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define RREPS(i,x) for(int i=((int)(x));i>0;i--)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();i++)
#define RFOR(i,c) for(__typeof((c).rbegin())i=(c).rbegin();i!=(c).rend();i++)
#define ALL(container) (container).begin(), (container).end()
#define RALL(container) (container).rbegin(), (container).rend()
#define SZ(container) ((int)container.size())
#define mp(a,b) make_pair(a, b)
#define pb push_back
#define eb emplace_back
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define __builtin_popcount __builtin_popcountll

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
    os<<"["; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"]"; return os;
}
template<class T> ostream& operator<<(ostream &os, const set<T> &t) {
    os<<"{"; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"}"; return os;
}
template<class S, class T> ostream& operator<<(ostream &os, const map<S, T> &t) {
    os<<"{"; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"}"; return os;
}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class S, class T> pair<S,T> operator+(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first+t.first, s.second+t.second);}
template<class S, class T> pair<S,T> operator-(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first-t.first, s.second-t.second);}

const int INF = 1<<28;
const double EPS = 1e-8;
const int MOD = 1000000007;


int T, h, w;
string s;
ll v[1001][1001];
ll vsum[1001][1001];

ll sum(int sx, int sy, int gx, int gy) {
    sx = max(min(sx, 1000), 0);
    sy = max(min(sy, 1000), 0);
    gx = max(min(gx, 1000), 0);
    gy = max(min(gy, 1000), 0);
    return vsum[gy][gx] - vsum[sy][gx] - vsum[gy][sx] + vsum[sy][sx];
}

ll solve(int l) {
    ll ans = 0;
    REP(i, h + w + 5) {
        REP(j, h + w + 5) {
            if (v[i][j]) {
                if (i + l <= h + w + 5 && v[i + l][j]) {
                    if (j + l <= h + w + 5) {
                        ans += sum(j + l, i, j + l + 1, i + l + 1);
                    }
                    if (j - l > 0) {
                        ans += sum(j - l, i, j - l + 1, i + l + 1);
                    }
                }
                if (j + l <= h + w + 5 && v[i][j + l]) {
                    if (i + l <= h + w + 5) {
                        ans += sum(j + 1, i + l, j + l, i + l + 1);
                    }
                    if (i - l > 0) {
                        ans += sum(j + 1, i - l, j + l, i - l + 1);
                    }
                }
            }
        }
    }
    return ans;
}

int main(int argc, char *argv[]){
    ios::sync_with_stdio(false);
    cin >> h >> w;
    REP(i, h) {
        string s;
        cin >> s;
        REP(j, w) {
            if (s[j] == '#') {
                v[i + j][h + j - i - 1] = 1;
            }
        }
    }
    REPS(i, 1000) {
        REPS(j, 1000) {
            vsum[i][j] = v[i - 1][j - 1] + vsum[i][j-1] + vsum[i-1][j] - vsum[i-1][j-1];
        }
    }
    ll ans = 0;
    REPS(l, 1000) {
        ans += solve(l);
    }
    cout << ans << endl;
    return 0;
}