CF1455B Jumps

You are standing on the $ \mathit{OX} $ -axis at point $ 0 $ and you want to move to an integer point $ x > 0 $ . You can make several jumps. Suppose you're currently at point $ y $ ( $ y $ may be negative) and jump for the $ k $ -th time. You can: - either jump to the point $ y + k $ - or jump to the point $ y - 1 $ . What is the minimum number of jumps you need to reach the point $ x $ ?

The first line contains a single integer $ t $ ( $ 1 \le t \le 1000 $ ) — the number of test cases. The first and only line of each test case contains the single integer $ x $ ( $ 1 \le x \le 10^6 $ ) — the destination point.

For each test case, print the single integer — the minimum number of jumps to reach $ x $ . It can be proved that we can reach any integer point $ x $ .

In the first test case $ x = 1 $ , so you need only one jump: the $ 1 $ -st jump from $ 0 $ to $ 0 + 1 = 1 $ . In the second test case $ x = 2 $ . You need at least three jumps: - the $ 1 $ -st jump from $ 0 $ to $ 0 + 1 = 1 $ ; - the $ 2 $ -nd jump from $ 1 $ to $ 1 + 2 = 3 $ ; - the $ 3 $ -rd jump from $ 3 $ to $ 3 - 1 = 2 $ ; Two jumps are not enough because these are the only possible variants: - the $ 1 $ -st jump as $ -1 $ and the $ 2 $ -nd one as $ -1 $ — you'll reach $ 0 -1 -1 =-2 $ ; - the $ 1 $ -st jump as $ -1 $ and the $ 2 $ -nd one as $ +2 $ — you'll reach $ 0 -1 +2 = 1 $ ; - the $ 1 $ -st jump as $ +1 $ and the $ 2 $ -nd one as $ -1 $ — you'll reach $ 0 +1 -1 = 0 $ ; - the $ 1 $ -st jump as $ +1 $ and the $ 2 $ -nd one as $ +2 $ — you'll reach $ 0 +1 +2 = 3 $ ; In the third test case, you need two jumps: the $ 1 $ -st one as $ +1 $ and the $ 2 $ -nd one as $ +2 $ , so $ 0 + 1 + 2 = 3 $ . In the fourth test case, you need three jumps: the $ 1 $ -st one as $ -1 $ , the $ 2 $ -nd one as $ +2 $ and the $ 3 $ -rd one as $ +3 $ , so $ 0 - 1 + 2 + 3 = 4 $ .