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>
using namespace std;

int sumh[601][601], sumw[601][601];

int main(){
    int H, W;
    cin >> H >> W;
    string S[300];
    for(int i=0; i<H; i++) cin >> S[i];

    bitset<601> S2[601];
    for(int i=0; i<H; i++){
        for(int j=0; j<W; j++){
            if(S[i][j] == '#') S2[i+j+1][i-j+W+1] = 1;
        }
    }

    for(int i=1; i<=600; i++){
        for(int j=1; j<=600; j++){
            sumh[i][j] = sumh[i-1][j] + S2[i][j];
            sumw[i][j] = sumw[i][j-1] + S2[i][j];
        }
    }

    int64_t ans = 0;
    for(int k=2; k<=600; k+=2){
        for(int i=1; i<=600; i++){
            for(int j=(i+W)%2; j<=600; j+=2){
                if(S2[i][j] == 0) continue;
                if(i+k <= 600 && S2[i+k][j]){
                    if(j-k >= 1)   ans += sumh[i+k-1][j-k] - sumh[i][j-k];
                    if(j+k <= 600) ans += sumh[i+k-1][j+k] - sumh[i][j+k];
                }
                if(j+k <= 600 && S2[i][j+k]){
                    if(i-k >= 1)   ans += sumw[i-k][j+k] - sumw[i-k][j-1];
                    if(i+k <= 600) ans += sumw[i+k][j+k] - sumw[i+k][j-1];
                }
            }
        }
    }
    cout << ans << endl;
    return 0;
}