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 nn can be a secret phone number. Only those permutations that minimize secret value might be the phone of her husband.

The sequence of 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 ai1,ai2,…,aikai1,ai2,…,aik where 1≤i1<i2<…<ik≤n1≤i1<i2<…<ik≤n is called increasing if ai1<ai2<ai3<…<aikai1<ai2<ai3<…<aik. If ai1>ai2>ai3>…>aikai1>ai2>ai3>…>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 nn (1≤n≤1051≤n≤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.

这是一道猜规律的题。开根号分块，我这样的菜鸡自然是猜不到的。。

不过还是可以证明的，面积一定的时候，周长最小的是正方形，其实也就是基本不等式嘛。。。

所以就是如果采用分块思想，可以得知，LIS是L，LDS则为ceil(n/L)，要使两者相加最小，也就是开根号的时候。

挺好的题，开阔一下思路

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define maxn 100
 #define maxm 30000
 #define ll long long
 #define mod 1000000007
 #define mem(a,b) memset(a,b,sizeof a)
 #ifndef ONLINE_JUDGE
    #define dbg(x) cout<<#x<<"="<<x<<endl;
 #else
    #define dbg(x)
 #endif
 inline int read()
 {
     int x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-') f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=*x+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 inline void Out(int a)
 {
     if(a>)
         Out(a/);
     putchar(a%+'');
 }
 int a[],b[];
 int main()
 {
      int n;
      while(~scanf("%d",&n)){
      for(int i=;i<=n;++i) a[i]=i;
      int hh=sqrt(n);
      int tot=n;
      int tt=n/hh;
      for(int i=;i<=tt;++i)
       {
             for(int j=;j<=hh;++j)
              {
                    cout<<a[tot-hh+j]<<" ";

              }
              tot-=hh;
       }
       for(int i=;i<=tot;++i) cout<<a[i]<<" ";
       cout<<endl;
      }
 }