B. Jumps
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 $$$x$$$?

Input

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.

Output

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 $$$x$$$.

Example
Input
5
1
2
3
4
5
Output
1
3
2
3
4
Note

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 $$$0 - 1 + 2 + 3 = 4$$$.