Problem:
Chain Disappearance Puzzle

We are playing a puzzle.
An upright board with H rows by 5 columns of cells, as shown in
the figure below, is used in this puzzle.
A stone engraved with a digit, one of 1 through 9, is placed in each of
the cells.
When three or more stones in horizontally adjacent cells are engraved
with the same digit, the stones will disappear.
If there are stones in the cells above the cell with a disappeared
stone, the stones in the above cells will drop down, filling the
vacancy.

The puzzle proceeds taking the following steps.

When three or more stones in horizontally adjacent cells are
engraved with the same digit, the stones will disappear. Disappearances
of all such groups of stones take place simultaneously.

When stones are in the cells above the emptied cells, these stones drop down so that the emptied cells are filled.

After the completion of all stone drops, if one or more groups of
stones satisfy the disappearance condition, repeat by returning to the
step 1.

The score of this puzzle is the sum of the digits on the disappeared stones.

Write a program that calculates the score of given configurations of stones.

Input

The input consists of multiple datasets.
Each dataset is formed as follows.

Board height H

Stone placement of the row 1

Stone placement of the row 2

...

Stone placement of the row H

The first line specifies the height ( H ) of the puzzle board (1 ≤ H ≤ 10).
The remaining H lines give placement of stones on each of the rows from top to bottom.
The placements are given by five digits (1 through 9), separated by a space.
These digits are engraved on the five stones in the corresponding row, in the same order.

The input ends with a line with a single zero.

Output

For each dataset, output the score in a line.
Output lines may not include any characters except the digits expressing the scores.

Sample Input

1
6 9 9 9 9
5
5 9 5 5 9
5 5 6 9 9
4 6 3 6 9
3 3 2 9 9
2 2 1 1 1
10
3 5 6 5 6
2 2 2 8 3
6 2 5 9 2
7 7 7 6 1
4 6 6 4 9
8 9 1 1 8
5 6 1 8 1
6 8 2 1 2
9 6 3 3 5
5 3 8 8 8
5
1 2 3 4 5
6 7 8 9 1
2 3 4 5 6
7 8 9 1 2
3 4 5 6 7
3
2 2 8 7 4
6 5 7 7 7
8 8 9 9 9
0

Output for the Sample Input

36
38
99
0
72

Solution in Ruby (judge verdict: Accepted)

module Graphical
	require 'set'

	def reachable_from(s, max_depth = nil)
		reached = Set[s]
		queue = [s]
		depth = 0
		loop {
			break if max_depth && depth >= max_depth
			next_queue = []
			until queue.empty?
				u = queue.shift
				each_out_connection_of(u){|v, _|
					next if reached.include?(v)
					reached << v
					next_queue << v
				}
			end
			break if next_queue.empty?
			queue = next_queue
			depth += 1
		}
		return reached
	end

	def bfs(start)
		raise "start must be Enumerable" if !start.kind_of?(Enumerable)
		reached = Set[*start.to_a]
		queue = []
		start.each{|s| queue << s}
		dist = {}
		loop.with_index {|_, depth|
			next_queue = []
			until queue.empty?
				u = queue.shift
				return depth if block_given? && yield(u, depth)
				dist[u] = depth
				each_out_connection_of(u){|v, _|
					next if reached.include?(v)
					reached << v
					next_queue << v
				}
			end
			break if next_queue.empty?
			queue = next_queue
		}
		return dist
	end

	def dijkstra(start)
		raise "start must be Enumerable" if !start.kind_of?(Enumerable)
		h = Heap.new
		reached = Set[*start.to_a]
		start.each{|s| h.add(s, 0)}

		dist = {}
		until h.empty?
			u, key = h.del_min
			return key if block_given? && yield(u, key)
			dist[u] = key
			each_out_connection_of(u){|v, l|
				d = dist[u] + l
				if reached.include?(v)
					h.update_key(v, d)
				else
					h.add(v, d)
					reached << v
				end
			}
		end
		block_given? ? nil : dist
	end

	def bidir_dijkstra(s, t)
		f_dist = {}
		b_dist = {}

		f_heap = Heap.new
		b_heap = Heap.new

		f_heap.add(s, 0)
		b_heap.add(t, 0)

		f_reached = Set[s]
		b_reached = Set[t]

		f_current_dist = b_current_dist = 0
		best = Float::INFINITY

		loop {
			if !f_heap.empty?
				u, key = f_heap.del_min
				f_dist[u] = key
				f_current_dist = key
				each_out_connection_of(u){|v, l|
					d = f_dist[u] + l
					if b_dist[v]
						path_len = d + b_dist[v]
						best = path_len if path_len < best
					end
					if f_reached.include?(v)
						f_heap.update_key(v, d)
					else
						f_heap.add(v, d)
						f_reached << v
					end
				}
			end

			if !b_heap.empty?
				u, key = b_heap.del_min
				b_dist[u] = key
				b_current_dist = key
				each_in_connection_of(u){|v, l|
					d = b_dist[u] + l
					if f_dist[v]
						path_len = d + f_dist[v]
						best = path_len if path_len < best
					end
					if b_reached.include?(v)
						b_heap.update_key(v, d)
					else
						b_heap.add(v, d)
						b_reached << v
					end
				}
			end

			return best if f_current_dist + b_current_dist >= best
			return nil if f_heap.empty? && b_heap.empty?
		}
	end

	def a_star(start)
		raise "a_star need termination condition" unless block_given?
		h = Heap.new
		reached = Set[start]
		h.add(start, [a_star_heuristic(start), 0])

		dist = {}
		until h.empty?
			u, (e, d) = h.del_min
			return d if yield(u, dist)
			dist[u] = d
			each_out_connection_of(u){|v, l|
				d = dist[u] + l
				e = d + a_star_heuristic(v)
				if reached.include?(v)
					h.update_key(v, [e, d])
				else
					h.add(v, [e, d])
					reached << v
				end
			}
		end
		block_given? ? nil : dist
	end

	#make_variations :bidir_dijkstra

	def prim(start)
		h = Heap.new
		h.add(start, 0)
		reached = Set[start]
		tree_len = 0

		until h.empty?
			u, key = h.del_min
			tree_len += key

			each_connection_of(u){|v, d|
				if reached.include?(v)
					h.update_key(v, d)
				else
					h.add(v, d)
					reached << v
				end
			}
		end
		tree_len
	end

	def each_layer_from(start)
		raise "start must be Enumerable" if !start.kind_of?(Enumerable)
		return enum_for(__method__, start) unless block_given?
		reached = Set[*start]
		current_layer = start

		loop {
			yield current_layer
			next_layer = []
			current_layer.each{|u|
				each_out_connection_of(u){|v|
					next if reached.include?(v)
					reached << v
					next_layer << v
				}
			}	

			return if next_layer.empty?
			current_layer = next_layer
		}
	end
end

module Grid
	def each_neighbour(u)
		y, x = u
		[[-1,0],[1,0],[0,-1],[0,1]].each{|dy, dx|
			i, j = y+dy, x+dx
			yield [i, j] if (0..self.size-1) === i && (0..self[0].size-1) === j
		}
	end
end

loop {
	h = gets.to_i
	break if h == 0
	$cols = (1..5).map{ [] }
	(0..h-1).each{|i|
		gets.split.map(&:to_i).each_with_index{|x, j|
			$cols[j][h-1-i] = x
		}
	}
	
	extend Graphical
	$cols.extend Grid

	def each_out_connection_of(u)
		x1, y1 = u
		$cols.each_neighbour(u){|v|
			x2, y2 = v
			yield(v, 1) if y2 == y1 && $cols[x2][y2] == $cols[x1][y1]
		}
	end

	score = 0
	
	loop {	
		prev_score = score
		(0..4).each{|j|
			(0..h-1).each{|i|
				if $cols[j][i]
					r = reachable_from([j, i])
					next if r.size < 3
					r.each{|x, y|
						score += $cols[x][y]
						$cols[x][y] = nil	
					}
				end
			}
		}
		break if prev_score == score
		$cols.map!(&:compact)
	}
	p score
}