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 <vector>
#include <string>

using namespace std;

vector<string> rotate45(const vector<string>& vs){
    const int H = vs.size();
    const int W = vs[0].size();
    vector<string> res(H+W-1, string(H+W-1, '.'));
    for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            res[i-j+W-1][i+j] = vs[i][j];
        }
    }
    return res;
}

vector<string> flip(const vector<string>& vs){
    const int H = vs.size();
    const int W = vs[0].size();
    vector<string> res(W, string(H, '.'));
    for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            res[j][i] = vs[i][j];
        }
    }
    return res;
}

pair<long long, long long> solve(const vector<string>& vs){
    const int H = vs.size();
    const int W = vs[0].size();
    vector<vector<long long>> sumH(H, vector<long long>(W+1, 0));
    for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            sumH[i][j+1] = sumH[i][j] + (vs[i][j] == '#' ? 1 : 0);
        }
    }
    long long res = 0;
    long long corner = 0;
    for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            if(vs[i][j] != '#') continue;
            for(int d=2;j+d<H;d+=2){
                if(vs[i][j+d] != '#') continue;
                for(int sgn=-1;sgn<=1;sgn+=2){
                    int ni = i + sgn * d;
                    if(0 <= ni && ni < H){
                        res += sumH[ni][j+d+1] - sumH[ni][j];
                        if(vs[ni][j] == '#') ++corner;
                        if(vs[ni][j+d] == '#') ++corner;
                    }
                }
            }
        }
    }
    return make_pair(res, corner);
}

int main(){
    int H, W;
    while(cin >> H >> W){
        vector<string> vs(H);
        for(auto& s : vs) cin >> s;
        auto board = rotate45(vs);
        auto r1 = solve(board);
        auto r2 = solve(flip(board));
        cout << (r1.first + r2.first - (r1.second + r2.second)/2) << endl;
    }
}