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;
typedef long long ll;
const int N=605;
int n,m,a[N][N];
ll ans;
char s[N][N];
int getsum(int x,int y)
{
    if(x<1||y<1) return 0;
    return a[x][y];
}
int flag;
void solve(int x,int y)
{
    while(x<=n&&y<=m)
    {
        if(s[x][y]=='#')
        for(int d=1;x+d<=n&&y+d<=m;d++)
            if(s[x+d][y+d]=='#')
        {
            ans+=getsum(x+2*d-flag,y-flag)-getsum(x+d-1+flag,y-d-1+flag);
            ans+=getsum(x-flag,y+2*d-flag)-getsum(x-d-1+flag,y+d-1+flag);
        }
        x++;y++;
    }
}
void solve()
{
    memset(a,0,sizeof(a));
    for(int i=1;i<=n*2;i++)
        for(int j=1;j<=m*2;j++)
        a[i][j]=a[i-1][j-1]+(s[i][j]=='#');
    for(int i=1;i<=n;i++)
        solve(i,1);
    for(int j=2;j<=m;j++)
        solve(1,j);
}
void rev()
{
    for(int i=1;i<=n;i++)
    {
        int l=1,r=m;
        while(l<=r) swap(s[i][l++],s[i][r--]);
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
    solve();
    rev();
    flag=1;
    solve();
    printf("%lld\n",ans);
}