Mrs. Smith is trying to contact her husband, John Smith, but she forgot the secret phone number!

The only thing Mrs. Smith remembered was that any permutation of $n$ can be a secret phone number. Only those permutations that minimize secret value might be the phone of her husband.

The sequence of $n$ integers is called a permutation if it contains all integers from $1$ to $n$ exactly once.

The secret value of a phone number is defined as the sum of the length of the longest increasing subsequence (LIS) and length of the longest decreasing subsequence (LDS). 

A subsequence $a_{i_1}, a_{i_2}, \ldots, a_{i_k}$ where $1\leq i_1 < i_2 < \ldots < i_k\leq n$ is called increasing if $a_{i_1} < a_{i_2} < a_{i_3} < \ldots < a_{i_k}$. If $a_{i_1} > a_{i_2} > a_{i_3} > \ldots > a_{i_k}$, a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.

For example, if there is a permutation $[6, 4, 1, 7, 2, 3, 5]$, LIS of this permutation will be $[1, 2, 3, 5]$, so the length of LIS is equal to $4$. LDS can be $[6, 4, 1]$, $[6, 4, 2]$, or $[6, 4, 3]$, so the length of LDS is $3$.

Note, the lengths of LIS and LDS can be different.

So please help Mrs. Smith to find a permutation that gives a minimum sum of lengths of LIS and LDS.


-----Input-----

The only line contains one integer $n$ ($1 \le n \le 10^5$) — the length of permutation that you need to build.


-----Output-----

Print a permutation that gives a minimum sum of lengths of LIS and LDS. 

If there are multiple answers, print any.


-----Examples-----
Input
4

Output
3 4 1 2

Input
2

Output
2 1



-----Note-----

In the first sample, you can build a permutation $[3, 4, 1, 2]$. LIS is $[3, 4]$ (or $[1, 2]$), so the length of LIS is equal to $2$. LDS can be ony of $[3, 1]$, $[4, 2]$, $[3, 2]$, or $[4, 1]$. The length of LDS is also equal to $2$. The sum is equal to $4$. Note that $[3, 4, 1, 2]$ is not the only permutation that is valid.

In the second sample, you can build a permutation $[2, 1]$. LIS is $[1]$ (or $[2]$), so the length of LIS is equal to $1$. LDS is $[2, 1]$, so the length of LDS is equal to $2$. The sum is equal to $3$. Note that permutation $[1, 2]$ is also valid.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'from math import sqrt\nn = int(input())\nr = int(sqrt(n))\nans = []\nif n == 1:\n\tprint("1")\nelif n == 2:\n\tprint("1 2")\nelif n == 3:\n\tprint("1 3 2")\nelif r ** 2 == n:\n\tfor i in range(r):\n\t\tans += [str((r - i - 1) * r + j) for j in range(1, r + 1)]\n\tprint(" ".join(ans))\nelse:\n\tif n <= r * (r + 1):\n\t\tsegment = r\n\t\telement = r + 1\n\telse:\n\t\tsegment = r + 1\n\t\telement = r + 1\n\tfor i in range(segment - 1):\n\t\tans += [str(n - element * i - j) for j in range(element - 1, -1, -1)]\n\tans += [str(j) for j in range(1, n - element * (segment - 1) + 1)]\n\tprint(" ".join(ans))'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13407