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}, …, a_{i_k} where 1≤ i_1 < i_2 < … < i_k≤ n is called increasing if a_{i_1} < a_{i_2} < a_{i_3} < … < a_{i_k}. If a_{i_1} > a_{i_2} > a_{i_3} > … > 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 ≤ n ≤ 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.
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class CF1017_C {
    public static void main(String[] args)throws Throwable {
        MyScanner sc=new MyScanner();
        PrintWriter pw=new PrintWriter(System.out);

        int n=sc.nextInt();
        int mn=n+1;
        int div=n,rep=1;
        for(int i=1;i<=n;i++){
            int curRep=(n+i-1)/i;
            int cnt=i+curRep;
            if(cnt<mn){
                mn=cnt;
                div=i;
                rep=curRep;
            }
        }
//        System.err.println(mn+" "+div+" "+rep);
        StringBuilder ans=new StringBuilder();
        int cur=n;
        for(int i=0;i<rep;i++){
            Stack<Integer> stack=new Stack<>();
            for(int j=0;j<div;j++){
                if(cur<1)
                    break;
                stack.add(cur);
                cur--;
            }
            while (!stack.isEmpty())
                ans.append(stack.pop()+" ");
        }
        pw.println(ans);
        pw.flush();
        pw.close();
    }
    static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {while (st == null || !st.hasMoreElements()) {
            try {st = new StringTokenizer(br.readLine());}
            catch (IOException e) {e.printStackTrace();}}
            return st.nextToken();}
        int nextInt() {return Integer.parseInt(next());}
        long nextLong() {return Long.parseLong(next());}
        double nextDouble() {return Double.parseDouble(next());}
        String nextLine(){String str = "";
            try {str = br.readLine();}
            catch (IOException e) {e.printStackTrace();}
            return str;}
    }
}