Problem:
Score: 400 points

Problem Statement

AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.

The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.

How many different PIN codes can he set this way?

Both the lucky number and the PIN code may begin with a 0.

Constraints

4 \leq N \leq 30000

S is a string of length N consisting of digits.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of different PIN codes Takahashi can set.

Sample Input 1

4
0224

Sample Output 1

3

Takahashi has the following options:

Erase the first digit of S and set 224.

Erase the second digit of S and set 024.

Erase the third digit of S and set 024.

Erase the fourth digit of S and set 022.

Thus, he can set three different PIN codes: 022, 024, and 224.

Sample Input 2

6
123123

Sample Output 2

17

Sample Input 3

19
3141592653589793238

Sample Output 3

329

Solution in Lua (5.3.2) (judge verdict: Accepted)

local function comp(a, b)
  return a < b
end

local function lower_bound(ary, x)
  local num = #ary
  if num == 0 then return 1 end
  if not comp(ary[1], x) then return 1 end
  if comp(ary[num], x) then return num + 1 end
  local min, max = 1, num
  while 1 < max - min do
    local mid = (min + max) // 2
    if comp(ary[mid], x) then
      min = mid
    else
      max = mid
    end
  end
  return max
end


local n = io.read("*n", "*l")
local s = io.read()
local t = {}
for i = 1, 10 do
  t[i] = {}
end
for i = 1, n do
  local v = s:sub(i, i):byte() - 48
  if v == 0 then v = 10 end
  table.insert(t[v], i)
end
local ret = 0
for i = 1, 10 do
  local lb1 = lower_bound(t[i], 1)
  if lb1 <= #t[i] then
    local p1 = t[i][lb1]
    for j = 1, 10 do
      local lb2 = lower_bound(t[j], p1 + 1)
      if lb2 <= #t[j] then
        local p2 = t[j][lb2]
        for k = 1, 10 do
          local lb3 = lower_bound(t[k], p2 + 1)
          if lb3 <= #t[k] then
            ret = ret + 1
          end
        end
      end
    end
  end
end
print(ret)