题目：
You are standing on the 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≤t≤1000) — the number of test cases.

The first and only line of each test case contains the single integer x (1≤x≤106) — 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
inputCopy
5
1
2
3
4
5
outputCopy
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.

#include <bits/stdc++.h>
using namespace std;
int main()
{
  int t;
  cin>>t;
  while(t--){
    long long n;
    scanf("%lld",&n);
    long long sum=0;
    int count=0;
    int i;
    for(i=0;i<=n;i++){
      sum+=i;
      if(sum>=n) break;
    }
    count=i;
    int ret=0;
    if(sum!=n){
      int b=abs(n-sum);
      if(b==1) count++;
      else count=count;
    }
    cout<<count<<endl;
  }
  return 0;
}