CF1455B.Jumps

You are standing on the $\mathit{OX}$-axis at point $0$ and you want to move to an integer point $x \gt 0$.

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 $x$?

你站在 $\mathit{OX}$-轴上的点 $0$ 处，希望移动到一个正整数点 $x > 0$。

你可以进行若干次跳跃。假设你当前位于点 $y$（$y$ 可能为负数），且这是第 $k$ 次跳跃。那么你可以：

• 跳跃至点 $y + k$；
• 或跳跃至点 $y - 1$。

到达点 $x$ 所需的最少跳跃次数是多少？

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.

第一行包含一个整数 $t$（$1 \le t \le 1000$）—— 测试用例的数量。

每个测试用例仅有一行，包含一个整数 $x$（$1 \le x \le 10^6$）—— 目标点。

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 $x$.

对于每个测试用例，输出一个整数——到达 $x$ 所需的最少跳跃次数。可以证明，我们总能到达任意整数点 $x$。

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

在第一个测试用例中，$x = 1$，因此你只需要一次跳跃：第 $1$ 次跳跃从 $0$ 到 $0 + 1 = 1$。

在第二个测试用例中，$x = 2$。你至少需要三次跳跃：

• 第 $1$ 次跳跃从 $0$ 到 $0 + 1 = 1$；
• 第 $2$ 次跳跃从 $1$ 到 $1 + 2 = 3$；
• 第 $3$ 次跳跃从 $3$ 到 $3 - 1 = 2$；

两次跳跃是不够的，因为仅存在以下四种可能情形：

• 第 $1$ 次跳跃为 $-1$，第 $2$ 次跳跃为 $-1$ —— 你将到达 $0 -1 -1 =-2$；
• 第 $1$ 次跳跃为 $-1$，第 $2$ 次跳跃为 $+2$ —— 你将到达 $0 -1 +2 = 1$；
• 第 $1$ 次跳跃为 $+1$，第 $2$ 次跳跃为 $-1$ —— 你将到达 $0 +1 -1 = 0$；
• 第 $1$ 次跳跃为 $+1$，第 $2$ 次跳跃为 $+2$ —— 你将到达 $0 +1 +2 = 3$；

在第三个测试用例中，你需要两次跳跃：第 $1$ 次为 $+1$，第 $2$ 次为 $+2$，即 $0 + 1 + 2 = 3$。

在第四个测试用例中，你需要三次跳跃：第 $1$ 次为 $-1$，第 $2$ 次为 $+2$，第 $3$ 次为 $+3$，即 $0 - 1 + 2 + 3 = 4$。