Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

5 4
#.##
.##.
#...
..##
...#

Sample Output 1

3

((1,1),(1,3),(2,2)),((1,1),(2,2),(3,1)) and ((1,3),(3,1),(4,4)) satisfy the condition.

Sample Input 2

13 27
......#.........#.......#..
#############...#.....###..
..............#####...##...
...#######......#...#######
...#.....#.....###...#...#.
...#######....#.#.#.#.###.#
..............#.#.#...#.#..
#############.#.#.#...###..
#...........#...#...#######
#..#######..#...#...#.....#
#..#.....#..#...#...#.###.#
#..#######..#...#...#.#.#.#
#..........##...#...#.#####

Sample Output 2

870

Solution in Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val H, W = in.next().toInt
  val s = new Array[String](H)
  for (i <- 0 until H) {
    s(i) = in.next()
  }

  //
  def manhattan(p1: (Int, Int), p2: (Int, Int)): Int = {
    val (x1, y1) = p1
    val (x2, y2) = p2

    (x1 - x2).abs + (y1 - y2).abs
  }

  var pairs: List[(Int, Int)] = Nil

  for(y <- 0 until H; x <- 0 until W) {
    if(s(y)(x) == '#') {
      pairs = (x+1, y+1) :: pairs
    }
  }

  var ans = 0

  val ar = pairs.toArray
  val l = ar.length
  for(i <- 0 until l-1) {
    for(k <- i+1 until l) {
      val dist = manhattan(ar(i), ar(k))
      for(m <- k+1 until l) {
        if(manhattan(ar(i), ar(m)) == dist && manhattan(ar(k), ar(m)) == dist)
          ans += 1
      }
    }
  }
  println(ans)
}


class InputReader(stream: InputStream) {
  private val reader = new BufferedReader(new InputStreamReader(stream))
  private var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}