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<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
using namespace std;
#define REP(i,m,n) for(int i=(int)m ; i < (int) n ; ++i )
#define rep(i,n) REP(i,0,n)
typedef long long ll;
typedef pair<int,int> pint;
typedef pair<ll,int> pli;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;


bitset<300> bs[300],ret;
int cnt[300];
int main(){
    int h,w;
    cin>>h>>w;
    string s[h];
    rep(i,h)cin>>s[i];
    rep(i,h)rep(j,w)bs[i][j]=(s[i][j]=='#');
    ll ans=0;
    rep(i,h)REP(j,i+1,h)REP(k,j+1,h){
        int x=j-i,y=k-j;
        if(x>=y)ret=bs[i]&(bs[j]<<x)&(bs[k]>>(x-y));
        else ret=bs[i]&(bs[j]<<(2*y-x))&(bs[k]<<(y-x));
        ans+=ret.count();
        if(x>=y)ret=bs[i]&(bs[j]>>x)&(bs[k]<<(x-y));
        else ret=bs[i]&(bs[j]>>(2*y-x))&(bs[k]>>(y-x));
        ans+=ret.count();
    }
    rep(i,h){
        rep(j,w)for(int k=2;j+k<w;k+=2){
            if(bs[i][j]&bs[i][j+k]){
                if(i+k/2<h)ans+=bs[i+k/2][j+k/2];
                if(i-k/2>=0)ans+=bs[i-k/2][j+k/2];
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}