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.util.*;
import java.io.*;
		
		public class codeforces
		{	
			public static void main(String[] args)
			{
				InputReader in = new InputReader(System.in);
				PrintWriter pw = new PrintWriter(System.out);
				
				int n = in.nextInt();
				int seg1 = 0, seg2 = 0, mins = 0, smin = 0, min = Integer.MAX_VALUE, m = 0, r = 0;
				
				for(int i=0;i<n;i++)
				{
					seg1 = n/(i+1) + Math.min(1, n%(i+1));
					seg2 = i+1;
					
					if(min-seg1-seg2>0)
					{
						mins = seg1;
						smin = seg2;
						m = n%(i+1);
						r = Math.min(1, n%(i+1));
						min = seg1+seg2;
					}
				}
				
				n++;
				
				for(int i=0;i<smin;i++)
				{
					if(m>0)
					{
						n = n - mins;
						
						for(int j=0;j<mins;j++)
							pw.print(n+j+" ");	
					}
					else
					{
						n = n - mins + r;
						
						for(int j=0;j<mins-r;j++)
							pw.print(n+j+" ");
					}
					m--;
				}
				
				pw.flush();
				pw.close();
			}
			
			static class pair implements Comparable<pair>
			{
				Long x, y;
				pair(long x,long y)
				{
					this.x=x;
					this.y=y;
				}
				
				public int compareTo(pair o) {
					int result = x.compareTo(o.x);
					if(result==0)
						result = y.compareTo(o.y);
					
					return result;
				}  
				
				public String toString()
				{
					return x+" "+y;
				}
				
				public boolean equals(Object o)
				{
					if (o instanceof pair)
				    {
						pair p = (pair)o;
						return p.x == x && p.y == y ;
					}
					return false;
				}
				
				public int hashCode()
				{
					return new Long(x).hashCode()*31 + new Long(y).hashCode();
				}
			}
			
			static class InputReader 
			{
				 
				private final InputStream stream;
				private final byte[] buf = new byte[8192];
				private int curChar, snumChars;
				private SpaceCharFilter filter;
		 
				public InputReader(InputStream stream) 
				{
					this.stream = stream;
				}
		 
				public int snext() 
				{
					if (snumChars == -1)
						throw new InputMismatchException();
					if (curChar >= snumChars) 
					{
						curChar = 0;
						try 
						{
							snumChars = stream.read(buf);
						} 
						catch (IOException e) 
						{
							throw new InputMismatchException();
						}
						if (snumChars <= 0)
							return -1;
					}
					return buf[curChar++];
				}
		 
				public int nextInt() 
			    {
					int c = snext();
					while (isSpaceChar(c)) 
					{
						c = snext();
					}
					int sgn = 1;
					if (c == '-')
				    {
						sgn = -1;
						c = snext();
					}
					int res = 0;
					do 
					{
						if (c < '0' || c > '9')
							throw new InputMismatchException();
						res *= 10;
						res += c - '0';
						c = snext();
					} while (!isSpaceChar(c));
					return res * sgn;
				}
		 
				public long nextLong()
			    {
					int c = snext();
					while (isSpaceChar(c)) 
					{
						c = snext();
					}
					int sgn = 1;
					if (c == '-') 
					{
						sgn = -1;
						c = snext();
					}
					long res = 0;
					do 
					{
						if (c < '0' || c > '9')
							throw new InputMismatchException();
						res *= 10;
						res += c - '0';
						c = snext();
					} while (!isSpaceChar(c));
					return res * sgn;
				}
		 
				public int[] nextIntArray(int n) 
				{
					int a[] = new int[n];
					for (int i = 0; i < n; i++) 
					{
						a[i] = nextInt();
					}
					return a;
				}
		 
				public String readString()
			    {
					int c = snext();
					while (isSpaceChar(c)) 
					{
						c = snext();
					}
					StringBuilder res = new StringBuilder();
					do 
					{
						res.appendCodePoint(c);
						c = snext();
					} while (!isSpaceChar(c));
					return res.toString();
				}
		 
				public String nextLine() 
				{
					int c = snext();
					while (isSpaceChar(c))
						c = snext();
					StringBuilder res = new StringBuilder();
					do 
					{
						res.appendCodePoint(c);
						c = snext();
					} while (!isEndOfLine(c));
					return res.toString();
				}
		 
				public boolean isSpaceChar(int c) 
				{
					if (filter != null)
						return filter.isSpaceChar(c);
					return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
				}
		 
				private boolean isEndOfLine(int c) 
				{
					return c == '\n' || c == '\r' || c == -1;
				}
		 
				public interface SpaceCharFilter
			    {
					public boolean isSpaceChar(int ch);
				}
			}
			
			static class CodeX {
	        public static void sort(long arr[]) {
	            merge_sort(arr, 0, arr.length - 1);
	        }
	
	        private static void merge_sort(long A[], long start, long end) {
	            if (start < end) {
	                long mid = (start + end) / 2;
	                merge_sort(A, start, mid);
	                merge_sort(A, mid + 1, end);
	                merge(A, start, mid, end);
	            }
	
	        }
	
	        private static void merge(long A[], long start,long mid, long end) {
	            long p = start, q = mid + 1;
	            long Arr[] = new long[(int)(end - start + 1)];
	            long k = 0;
	
	            for (int i = (int)start; i <= end; i++) {
	                if (p > mid)
	                    Arr[(int)k++] = A[(int)q++];
	
	                else if (q > end)
	                    Arr[(int)k++] = A[(int)p++];
	
	                else if (A[(int)p] < A[(int)q])
	                    Arr[(int)k++] = A[(int)p++];
	
	                else
	                    Arr[(int)k++] = A[(int)q++];
	            }
	            for (int i = 0; i < k; i++) {
	                A[(int)start++] = Arr[i];
	            }
	
	        }
	
	    }
		}	
			