CF502C The Phone Number

C. The Phone Number

time limit per test

1 second

memory limit per test

256 megabytes

 

 

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 $[3,4,1,2][3,4,1,2].$

$LIS is [3,4][3,4] (or [1,2][1,2]),\; so\; the\; length\; of LIS is\; equal\; to 22.$

$LDS can\; be\; ony\; of [3,1][3,1], [4,2][4,2], [3,2][3,2],\; or [4,1][4,1].$

$The\; length\; of LDS is\; also\; equal\; to 22.$

$T$$he\; 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 $[2,1][2,1]. LIS is [1][1] (or [2][2]),$

$so\; the\; length\; of LIS is\; equal\; to 11.$

$LDS is [2,1][2,1],\; so\; the\; length\; of LDS is\; equal\; to 22.$

$The\; sum\; is\; equal\; to 33.$

$Note\; that\; permutation [1,2][1,2] is\; also\; valid.$

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <iomanip>
 4 #include <cstring>
 5 #include <climits>
 6 #include <complex>
 7 #include <fstream>
 8 #include <cassert>
 9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19 using namespace std;
20 #define INF 0x3f3f3f3f
21 #define maxn 100
22 #define maxm 30000
23 #define ll long long
24 #define mod 1000000007
25 #define mem(a,b) memset(a,b,sizeof a)
26 #ifndef ONLINE_JUDGE
27    #define dbg(x) cout<<#x<<"="<<x<<endl;
28 #else 
29    #define dbg(x) 
30 #endif
31 inline int read()
32 {
33     int x=0,f=1;
34     char ch=getchar();
35     while(ch<'0'||ch>'9')
36     {
37         if(ch=='-') f=-1;
38         ch=getchar();
39     }
40     while(ch>='0'&&ch<='9')
41     {
42         x=10*x+ch-'0';
43         ch=getchar();
44     }
45     return x*f;
46 }
47 inline void Out(int a) 
48 {
49     if(a>9)
50         Out(a/10);
51     putchar(a%10+'0');
52 }
53 int a[100005],b[100005];
54 int main()
55 {
56      int n;
57      while(~scanf("%d",&n)){
58      for(int i=1;i<=n;++i) a[i]=i;
59      int hh=sqrt(n);
60      int tot=n;
61      int tt=n/hh;
62      for(int i=1;i<=tt;++i)
63       {
64             for(int j=1;j<=hh;++j)
65              {
66                    cout<<a[tot-hh+j]<<" ";
67                    
68              }
69              tot-=hh;
70       }
71       for(int i=1;i<=tot;++i) cout<<a[i]<<" ";
72       cout<<endl;
73      }
74 }

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