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>
#define For(i, a, b) for(int (i)=(int)(a); (i)<(int)(b); ++(i))
#define rFor(i, a, b) for(int (i)=(int)(a)-1; (i)>=(int)(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef unsigned long long ulint;
typedef pair<int, int> pii;
template<class T> bool chmax(T &a, const T &b){if(a<b){a=b; return true;} return false;}
template<class T> bool chmin(T &a, const T &b){if(a>b){a=b; return true;} return false;}
template<class T> T div_floor(T a, T b){
    if(b < 0) a *= -1, b *= -1;
    return a>=0 ? a/b : (a+1)/b-1;
}
template<class T> T div_ceil(T a, T b){
    if(b < 0) a *= -1, b *= -1;
    return a>0 ? (a-1)/b+1 : a/b;
}

constexpr lint mod = 1e9+7;
constexpr lint INF = mod * mod;
constexpr int MAX = 200010;

int H, W;
char c[610][610];

void rot(){
    char t[W][H];
    rep(i, H)rep(j, W) t[j][i] = c[i][j];
    swap(H, W);
    rep(i, H)rep(j, W) c[i][j] = t[i][j];
}

lint f(int flag){
    lint ret = 0;
    int cnt[H][W+1];
    rep(i, H){
        cnt[i][0] = 0;
        rep(j, W) cnt[i][j+1] = cnt[i][j] + (c[i][j] == '#');
    }
    rep(i, H){
        rep(s, W-1)if(c[i][s] == '#')For(t, s+1, W)if(c[i][t] == '#'){
            int d = t - s;
            if(i-d >= 0) ret += cnt[i-d][t+1] - cnt[i-d][s];
            if(i+d < H) ret += cnt[i+d][t+1] - cnt[i+d][s];

            if(flag){
                if(i-d >= 0) ret -= (c[i-d][s] == '#') + (c[i-d][t] == '#');
                if(i+d < H) ret -= (c[i+d][s] == '#') + (c[i+d][t] == '#');
            }
        }
    }
    return ret;
}

int main(){
    int h, w;
    scanf("%d%d", &h, &w);
    H = h + w, W = 2 * max(h, w);
    rep(i, H)rep(j, W) c[i][j] = '.';
    rep(i, h)rep(j, w){
        char t;
        scanf(" %c", &t);
        c[i+j][i-j+W/2] = t;
    }
    lint ans = f(0);
    rot();
    ans += f(1);
    printf("%lld\n", ans);
}