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 Task {

    public static void main(String[] args) throws IOException {

        new Task().go();
    }

    PrintWriter out;
    Reader in;
    BufferedReader br;

    Task() throws IOException {

        try {

            //br = new BufferedReader( new FileReader("input.txt") );
            in = new Reader("input.txt");
            out = new PrintWriter( new BufferedWriter(new FileWriter("output.txt")) );
        }
        catch (Exception e) {

            //br = new BufferedReader( new InputStreamReader( System.in ) );
            in = new Reader();
            out = new PrintWriter( new BufferedWriter(new OutputStreamWriter(System.out)) );
        }
    }

    void go() throws IOException {

        int t = 1;
        while (t > 0) {
            solve();
            //out.println();
            t--;
        }

        out.flush();
        out.close();
    }

    int inf = 2000000000;
    int mod = 1000000007;
    double eps = 0.000000001;

    int n;
    int m;

    int[] a;
    ArrayList<Integer>[] g;
    void solve() throws IOException {

        int n = in.nextInt();
        int d = 1;
        while (d * d < n)
            d++;
        int cur = n;
        for (int i = 0; i < n;) {
            for (int j = cur - d + 1; j <= cur; j++)
                if (j > 0) {
                    out.print(j + " ");
                    i++;
                }
            cur -= d;
        }
    }

    int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    class Pair implements Comparable<Pair>{
        int a;
        int b;

        Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }

        public int compareTo(Pair p) {
            if (a > p.a) return 1;
            if (a < p.a) return -1;
            if (b > p.b) return 1;
            if (b < p.b) return -1;
            return 0;
        }
    }

    class Item {

        int a;
        int b;
        int c;

        Item(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

    }

    class Reader {

        BufferedReader  br;
        StringTokenizer tok;

        Reader(String file) throws IOException {
            br = new BufferedReader( new FileReader(file) );
        }

        Reader() throws IOException {
            br = new BufferedReader( new InputStreamReader(System.in) );
        }

        String next() throws IOException {

            while (tok == null || !tok.hasMoreElements())
                tok = new StringTokenizer(br.readLine());
            return tok.nextToken();
        }

        int nextInt() throws NumberFormatException, IOException {
            return Integer.valueOf(next());
        }

        long nextLong() throws NumberFormatException, IOException {
            return Long.valueOf(next());
        }

        double nextDouble() throws NumberFormatException, IOException {
            return Double.valueOf(next());
        }


        String nextLine() throws IOException {
            return br.readLine();
        }

        int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        ArrayList<Integer>[] nextGraph(int n, int m) throws IOException {
            ArrayList<Integer>[] g = new ArrayList[n];
            for (int i = 0; i < n; i++)
                g[i] = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                int x = nextInt() - 1;
                int y = nextInt() - 1;
                g[x].add(y);
                g[y].add(x);
            }

            return g;
        }

    }

}