Problem:
Max Score: 350 Points

Problem Statement

There are N buildings along the line. The i-th building from the left is colored in color i, and its height is currently a_i meters.

Chokudai is a mayor of the city, and he loves colorful thigs. And now he wants to see at least K buildings from the left.

You can increase height of buildings, but it costs 1 yens to increase 1 meters. It means you cannot make building that height is not integer.

You cannot decrease height of buildings.

Calculate the minimum cost of satisfying Chokudai's objective.

Note: "Building i can see from the left" means there are no j exists that (height of building j) ≥ (height of building i) and j < i.

Input Format

N K
a_1 a_2 a_3 ... a_N

Output Format

Print the minimum cost in one line. In the end put a line break.

Constraints

1 ≤ K ≤ N ≤ 15

1 ≤ a_i ≤ 10^9

Scoring

Subtask 1 [120 points]

N = K

Subtask 2 [90 points]

N ≤ 5

a_i ≤ 7

Subtask 3 [140 points]

There are no additional constraints.

Sample Input 1

5 5
3949 3774 3598 3469 3424

Sample Output 1

1541

The optimal solution is (height of buildings from the left) = [3949, 3950, 3951, 3952, 3953].

Sample Input 2

5 3
7 4 2 6 4

Sample Output 2

7

The optimal solution is (height of buildings from the left) = [7, 8, 2, 9, 4].

Solution in Lua (Lua 5.3.5) (judge verdict: Wrong Answer)

local mmi, mma = math.min, math.max
local function combination_get_map(n)
  local combmap = {}
  for i = 1, n do
    combmap[i] = {}
    local ret = 1
    for j = 1, i do
      ret = ret * (i + 1 - j) // j
      combmap[i][j] = ret
    end
  end
  return combmap
end
local function combination_get_count(combmap, n, k)
  if k < 0 then return 0
  elseif n < k then return 0
  elseif k == 0 then return 1
  else return combmap[n][k]
  end
end

local function combination_get_array(combmap, n, k, idx)
  local rem = k
  local retary = {}
  for i = 1, n do
    local useCnt = combination_get_count(combmap, n - i, rem - 1)
    local unuseCnt = combination_get_count(combmap, n - i, rem)
    if idx <= useCnt then
      rem = rem - 1
      table.insert(retary, i)
    else
      idx = idx - useCnt
    end
  end
  return retary
end

local n, k = io.read("*n", "*n")
local a = {}
for i = 1, n do
  a[i] = io.read("*n")
end
local cmap = combination_get_map(n)
local tot = combination_get_count(cmap, n, k)
local mincost = 1000000007 * k
for i = 1, tot do
  local cost = 0
  local ary = combination_get_array(cmap, n, k, i)
  local height = a[1]
  for j = 2, ary[1] do
    height = mma(height, a[j])
  end
  for j = 2, k do
    local v = a[ary[j]]
    if height < v then
      height = v
    else
      cost = cost + height + 1 - v
      height = height + 1
    end
  end
  mincost = mmi(mincost, cost)
end
print(mincost)