CodeForces - 1017 C. The Phone Number（数学）

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

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

题意：给定N，输出一个N的排列，满足其LIS+LDS最小。

思路：当时猜到了是根号倒序排序，但是写WA了。 应该是有什么证明的，不过不难想。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000010;
int a[maxn];
int main()
{
    int N,tN,i,j;
    scanf("%d",&N); tN=sqrt(N); int p=N;
    for(i=1;i<=N;i+=tN){
        for(j=min(N,i+tN-1);j>=i;j--) a[j]=p--;
    }
    for(i=1;i<=N;i++) printf("%d ",a[i]);
    return 0;
}