Problem:
Score : 200 points

Problem Statement

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0.
During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right.
That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i.
The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible.
Find the earliest possible time to reach coordinate X.

Constraints

X is an integer.

1≤X≤10^9

Input

The input is given from Standard Input in the following format:

X

Output

Print the earliest possible time for the kangaroo to reach coordinate X.

Sample Input 1

6

Sample Output 1

3

The kangaroo can reach his nest at time 3 by jumping to the right three times, which is the earliest possible time.

Sample Input 2

2

Sample Output 2

2

He can reach his nest at time 2 by staying at his position during the first second, and jumping to the right at the next second.

Sample Input 3

11

Sample Output 3

5

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

main = do
  li <- getLine
  let x = read li
  let ans = compute (abs x)
  print ans

{-
2に行くのに-3,+5とするよりは+2の方が早いように、
逆向きを交えて狙う必要は実はないのだと思われる。
つまり時刻nには1+...+n = n(n+1)/2まで進むことができて、
それより手前はいずれかを落とすことで必ず到達できる。（要証明）
差がn以下ならその数ズバリを落とせばいい。
なので、xをカバーする最小のnを計算で求める。
n(n+1)/2 >= x
n^2 + n - 2x >= 0
n = -1 +- sqrt(1 + 8x)/2
-}

compute :: Int -> Int
compute x = ceiling ((sqrt (fromIntegral x * 8 + 1) - 1) / 2)