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.util.*;
public class Main
{
	public static void main(String[] args) throws IOException
	{
		// int t = ni();
		int t = 1;
		while(t-->0) {
			long n = ni();
            int a = (int)Math.ceil(Math.sqrt(n));
            int b = (int)Math.floor(Math.sqrt(n));
            // System.out.println(a+" "+b);
            if(n==a*a) {
                int var = a;
                while(var-->0){
                    for(int i=var*a+1;i<=(var+1)*a;i++) {
                        ans.append(i+" ");
                    }
                }
            }
            else if(n<=a*b) {
                int var = b;
                while(var-->0){
                    for(int i=var*a+1;i<=(var+1)*a && i<=n;i++) {
                        ans.append(i+" ");
                    }
                }
            }
            else if(n>a*b) {
                int var = b;
                for(int i=a*b+1;i<=n;i++) {
                    ans.append(i+" ");
                }
                while(var-->0){
                    for(int i=var*a+1;i<=(var+1)*a && i<=n;i++) {
                        ans.append(i+" ");
                    }
                }
            }
		}
		System.out.println(ans);
	}

	static final int mod=1000000007;
    static StringBuilder ans=new StringBuilder();
	static Reader in=new Reader();
	static final double eps=1e-9;

	//////////////////////////////////
	//input functions/////////////////
	static int ni(){return Integer.parseInt(in.next());}
	static long nl(){return Long.parseLong(in.next());}
	static String ns(){return in.next();}
	static double nd(){return Double.parseDouble(in.next());}
	static int[] nia(int n){int a[]=new int[n];for(int i=0; i<n; i++)a[i]=ni();return a;}
	static int[] pnia(int n){int a[]=new int[n+1];for(int i=1; i<=n; i++)a[i]=ni();return a;}
	static long[] nla(int n){long a[]=new long[n];for(int i=0; i<n; i++)a[i]=nl();return a;}
	static String[] nsa(int n){String a[]=new String[n];for(int i=0; i<n; i++)a[i]=ns();return a;}
	static double[] nda(int n){double a[]=new double[n];for(int i=0; i<n; i++)a[i]=nd();return a;}

	//////output functions////////////////
	static void pr(Object a){ans.append(a+"\n");}
	static void pr(){ans.append("\n");}
	static void p(Object a){ans.append(a);}
	static void pra(int[]a){for(int i:a)ans.append(i+" ");ans.append("\n");}
	static void pra(long[]a){for(long i:a)ans.append(i+" ");ans.append("\n");}
	static void pra(String[]a){for(String i:a)ans.append(i+" ");ans.append("\n");}
	static void pra(double[]a){for(double i:a)ans.append(i+" ");ans.append("\n");}
	static void sop(Object a){System.out.println(a);}
	static void flush(){System.out.print(ans);ans=new StringBuilder();}

	static void exit(){System.out.print(ans);System.exit(0);}
	static int min(int... a){int min=a[0];for(int i:a)min=Math.min(min, i);return min;}
	static int max(int... a){int max=a[0];for(int i:a)max=Math.max(max, i);return max;}	
	static int gcd(int... a){int gcd=a[0];for(int i:a)gcd=gcd(gcd, i);return gcd;}	
	static long min(long... a){long min=a[0];for(long i:a)min=Math.min(min, i);return min;}
	static long max(long... a){long max=a[0];for(long i:a)max=Math.max(max, i);return max;}	
	static long gcd(long... a){long gcd=a[0];for(long i:a)gcd=gcd(gcd, i);return gcd;}	
	static String pr(String a, long b){String c="";while(b>0){if(b%2==1)c=c.concat(a);a=a.concat(a);b>>=1;}return c;}
	static long powm(long a, long b, long m){long an=1;long c=a;while(b>0){if(b%2==1)an=(an*c)%m;c=(c*c)%m;b>>=1;}return an;}
	static int gcd(int a, int b){if(b==0)return a;return gcd(b, a%b);}
	static long gcd(long a, long b){if(b==0)return a;return gcd(b, a%b);}

	static class Reader {
        BufferedReader reader;
        StringTokenizer tokenizer;
        public Reader() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        String next() {while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {tokenizer = new StringTokenizer(reader.readLine());}
				catch (IOException e) {throw new RuntimeException(e);}}
            return tokenizer.nextToken();}
        int nextInt() {return Integer.parseInt(next());}
        long nextLong() {return Long.parseLong(next());}
        double nextDouble() {return Double.parseDouble(next());}
    }

}
