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 lli;
typedef double lld;
typedef vector<lli> vll;
typedef vector<bool> vbl;
typedef vector<double> vdl;
typedef vector<vector<lli>> mat;
typedef vector<vdl> mad;
typedef unordered_map<lli,unordered_map<lli,lli>> graph;
typedef complex<double> cmp;
typedef vector<cmp> vcl;

lli H = 300*2+10;
lli W = 300*2+10;
lli h,w;
mat s;
lli ans = 0;

int main(){
  cin >> h >> w;
  s = mat(H, vll(W));
  for(lli i = 1;i <= h;i++){
    for(lli j = 1;j <= w;j++){
      char c;
      cin >> c;
      if(c == '#') s[i+j][i-j+300] = 1;
      else s[i+j][i-j+300] = 0;
    }
  }
  mat x = s;
  mat y = s;
  for(lli i = 0;i < H;i++) for(lli j = 1;j < W;j++) x[i][j] += x[i][j-1];
  for(lli i = 0;i < W;i++) for(lli j = 1;j < H;j++) y[j][i] += y[j-1][i];
  for(lli i = 0;i < H;i++){
    for(lli from = 0;from < W;from++){
      if(s[i][from] != 1) continue;
      for(lli to = from+1;to < W;to++){
        if(s[i][to] == 1){
          lli d = to-from;
          if(i-d >= 0){
            ans += x[i-d][to]-x[i-d][from-1];
          }
          if(i+d < H){
            ans += x[i+d][to]-x[i+d][from-1];
          }
        }
      }
    }
  }
  for(lli i = 0;i < W;i++){
    for(lli from = 0;from < H;from++){
      if(s[from][i] != 1) continue;
      for(lli to = from+1;to < H;to++){
        if(s[to][i] == 1){
          lli d = to-from;
          if(i-d >= 0){
            ans += y[to-1][i-d]-y[from][i-d];
          }
          if(i+d < W){
            ans += y[to-1][i+d]-y[from][i+d];
          }
        }
      }
    }
  }
  cout << ans << endl;
}