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.IOException; 
import java.io.InputStream; 
import java.io.PrintWriter;
import java.util.*;


public class c{ 

	static void solve(){ 
		int n = ni();
		int m =1;
		while(m*m<n)++m;
		Deque<Integer> que = new ArrayDeque<>();
		int base=0;
		List<Integer> ans = new ArrayList<>();
		while(base<n){
			for(int i=base+1;i<=Math.min(n,base+m);++i)que.add(i);
			while(!que.isEmpty())ans.add(que.removeLast());
			base+=m;
		}
		for(int i=0;i<n;++i)out.print(ans.get(i)+(i==n-1 ? "\n":" "));
 
	} 
 
 
 
 
	public static void main(String[] args){ 
		 solve(); 
		 out.flush(); 
	 } 
	 private static InputStream in = System.in; 
	 private static PrintWriter out = new PrintWriter(System.out); 
 
	 private static final byte[] buffer = new byte[1<<15]; 
	 private static int ptr = 0; 
	 private static int buflen = 0; 
	 private static boolean hasNextByte(){ 
		 if(ptr<buflen)return true; 
		 ptr = 0; 
		 try{ 
			 buflen = in.read(buffer); 
		 } catch (IOException e){ 
			 e.printStackTrace(); 
		 } 
		 return buflen>0; 
	 } 
	 private static int readByte(){ if(hasNextByte()) return buffer[ptr++]; else return -1;} 
	 private static boolean isSpaceChar(int c){ return !(33<=c && c<=126);} 
	 private static int skip(){int res; while((res=readByte())!=-1 && isSpaceChar(res)); return res;} 
 
	 private static double nd(){ return Double.parseDouble(ns()); } 
	 private static char nc(){ return (char)skip(); } 
	 private static String ns(){ 
		 StringBuilder sb = new StringBuilder(); 
		 for(int b=skip();!isSpaceChar(b);b=readByte())sb.append((char)b); 
		 return sb.toString(); 
	 } 
	 private static int[] nia(int n){ 
		 int[] res = new int[n]; 
		 for(int i=0;i<n;++i)res[i]=ni(); 
		 return res; 
	 } 
	 private static long[] nla(int n){ 
		 long[] res = new long[n]; 
		 for(int i=0;i<n;++i)res[i]=nl(); 
		 return res; 
	 } 
	 private static int ni(){ 
		 int res=0,b; 
		 boolean minus=false; 
		 while((b=readByte())!=-1 && !((b>='0'&&b<='9') || b=='-')); 
		 if(b=='-'){ 
			 minus=true; 
			 b=readByte(); 
		 } 
		 for(;'0'<=b&&b<='9';b=readByte())res=res*10+(b-'0'); 
		 return minus ? -res:res; 
	 } 
	 private static long nl(){ 
		 long res=0,b; 
		 boolean minus=false; 
		 while((b=readByte())!=-1 && !((b>='0'&&b<='9') || b=='-')); 
		 if(b=='-'){ 
			 minus=true; 
			 b=readByte(); 
		 } 
		 for(;'0'<=b&&b<='9';b=readByte())res=res*10+(b-'0'); 
		 return minus ? -res:res; 
	} 
} 

