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;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcountll

#define INF 1e16
#define mod 1000000007

ll h,w,n;
char s[700][700];
ll sumx[700][700],sumy[700][700];

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  cin>>h>>w;
  rep(i,h){
    string in;
    cin>>in;
    rep(j,w){
      s[i+j][i-j+w-1]=in[j];
    }
  }
  n=h+w;
  rep(i,n)rep(j,n)if(s[i][j]==0)s[i][j]='.';
  ll res=0;
  rep(i,n)rep(j,n){
    if(s[i][j]=='#'){
      sumx[i][j+1]=sumx[i][j]+1;
      sumy[i+1][j]=sumy[i][j]+1;
    }else{
      sumx[i][j+1]=sumx[i][j];
      sumy[i+1][j]=sumy[i][j];
    }
  }
  rep(i,n){
    rep(j,n)repl(k,j+1,n){
      if(s[i][j]=='#'&&s[i][k]=='#'){
        ll d=k-j;
        if(i-d>=0)res+=sumx[i-d][k+1]-sumx[i-d][j];
        if(i+d<n)res+=sumx[i+d][k+1]-sumx[i+d][j];
      }
    }
  }
  rep(j,n){
    rep(i,n)repl(k,i+1,n){
      if(s[i][j]=='#'&&s[k][j]=='#'){
        ll d=k-i;
        if(j-d>=0)res+=sumy[k][j-d]-sumy[i+1][j-d];
        if(j+d<n)res+=sumy[k][j+d]-sumy[i+1][j+d];
      }
    }
  }

  cout<<res<<endl;

  return 0;
}