cf 1017C

C. The Phone Number

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $\mathrm{nn can\; be\; a\; secret\; phone\; number.\; Only\; those\; permutations\; that\; minimize\; secret\; value\; might\; be\; the\; phone\; of\; her\; husband.}$

The sequence of $\mathrm{nn integers\; is\; called\; a\; permutation\; if\; it\; contains\; all\; integers\; from 11 to nn exactly\; once.}$

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 ${\mathrm{ai1,ai2,\dots ,aikai1,ai2,\dots ,aik where 1\le i1<i2<\dots <ik\le n1\le i1<i2<\dots <ik\le n is\; called\; increasing\; if ai1<ai2<ai3<\dots <aikai1<ai2<ai3<\dots <aik.\; If ai1>ai2>ai3>\dots >aikai1>ai2>ai3>\dots >aik,\; a\; subsequence\; is\; called\; decreasing.\; An\; increasing/decreasing\; subsequence\; is\; called\; longest\; if\; it\; has\; maximum\; length\; among\; all\; increasing/decreasing\; subsequences.}}^{}$

For example, if there is a permutation $[6,4,1,7,2,3,5][6,4,1,7,2,3,5], LIS of\; this\; permutation\; will\; be [1,2,3,5][1,2,3,5],\; so\; the\; length\; of LIS is\; equal\; to 44. LDS can\; be [6,4,1][6,4,1], [6,4,2][6,4,2],\; or [6,4,3][6,4,3],\; so\; the\; length\; of LDS is 33.$

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 $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; length\; of\; permutation\; that\; you\; need\; to\; build.}$

Output

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

If there are multiple answers, print any.

Examples

input

Copy

4

output

Copy

3 4 1 2

input

Copy

2

output

Copy

2 1

Note

In the first sample, you can build a permutation $[3,4,1,2][3,4,1,2]. LIS is [3,4][3,4] (or [1,2][1,2]),\; so\; the\; length\; of LIS is\; equal\; to 22. LDS can\; be\; ony\; of [3,1][3,1], [4,2][4,2], [3,2][3,2],\; or [4,1][4,1].\; The\; length\; of LDS is\; also\; equal\; to 22.\; The\; sum\; is\; equal\; to 44.\; Note\; that [3,4,1,2][3,4,1,2] is not the\; only\; permutation\; that\; is\; valid.$

In the second sample, you can build a permutation $[2,1][2,1]. LIS is [1][1] (or [2][2]),\; so\; the\; length\; of LIS is\; equal\; to 11. LDS is [2,1][2,1],\; so\; the\; length\; of LDS is\; equal\; to 22.\; The\; sum\; is\; equal\; to 33.\; Note\; that\; permutation [1,2][1,2] is\; also\; valid.$

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
#define  N 100009
#define ll long long 
#define  mem(a,b)  memset(a,b,sizeof(a))
int n,a[N];
/*
分成x组，每一组y个，x*y==n,求x+y的最小值，很明显x==sqrt(n)时成立
当n%sqrt(n)==0时，令m=n/sqrt(n)，即x*m==n
那么分成x组：LDS==x，每一组里的数升序，总体递减
每一组m个： LIS==m
每一组x个 ：LDS=x
分成y组  ： LIS=m
当n%sqrt!=0 时，就是再将剩下的放到另一组里
*/
int  main()
{
    scanf("%d",&n);
    int index=sqrt(n);//每一组index个
    int tmp=1,t;
    for(int  i=n-index;i>=0;i-=index){
        t=i;
        for(int j=0;j<index;j++){
            a[i+j]=tmp++;
        }
    }
    for(int i=0;i<t;i++)  a[i]=tmp++;//剩下的再放到另一个组里
    for(int i=0;i<n;i++){
        printf("%d%c",a[i],i==n-1?'\n':' ');
    }
    return 0;
}