A - Infinite Sequence

Problem description

Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).

Find the number on the n-th position of the sequence.

Input

The only line contains integer n (1 ≤ n ≤ 10¹⁴) — the position of the number to find.

Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use longinteger type.

Output

Print the element in the n-th position of the sequence (the elements are numerated from one).

Examples

Input

3

Output

2

Input

5

Output

2

Input

10

Output

4

Input

55

Output

10

Input

56

Output

1
解题思路：简单推导：假设k表示为第k(k>=1)个序列（1，...，k），那么前k(k>=1)个序列一共有n=（k+1）*k/2个数，则第n个数在第floor（√（2*n+0.25）-0.5）（向下取整）个序列中。因为当n必须刚好为前k个序列数字的总个数时，k才会是那个精确的第k个序列。因此，此时只需判断m=（k+1）*k/2是否大于n。如果n大于m，说明第n个数实际在第k+1个序列里面，则这个数为n-m；否则第n个数就在第k个序列里面，则这个数为k-(m-n)=k+n-m。
AC代码：

 #include<bits/stdc++.h>
 using namespace std;
 int main(){
     long long n,k,m;
     cin>>n;
     k=sqrt(2.0*n+0.25)-0.5;
     m=k*(k+)/;
     if(n>m)cout<<n-m<<endl;
     else cout<<k+n-m<<endl;
     return ;
 }