Solve the programming task below in a Python markdown code block.
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 $x$.


-----Examples-----

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