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.BufferedReader;
// import java.io.FileInputStream;
// import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

import static java.lang.Math.abs;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.round;
import static java.lang.Math.sqrt;
import static java.util.Arrays.copyOf;
import static java.util.Arrays.fill;
import static java.util.Arrays.sort;
import static java.util.Collections.reverse;
import static java.util.Collections.reverseOrder;
import static java.util.Collections.sort;
import static java.util.Comparator.comparing;
import static java.util.Comparator.comparingInt;
import static java.util.Comparator.comparingLong;

public class Main {
    FastScanner in;
    PrintWriter out;

    private void solve() throws IOException {
        solveC();
        // solveE();
        // solveF();
        // solveG();
        // solveH();
    }

    private void solveC() throws IOException {
        int n = in.nextInt();
        int[] a = new int[n];

        int s = (int) sqrt(n);
        while (s * s < n)
            s++;
        while (s * s > n)
            s--;

        int x = n;
        for (int i = n - s; i >= 0; i -= s) {
            for (int j = 0; j < s; j++) {
                a[i + j] = x--;
            }
        }


        for (int i = 0; i < n && x > 0; i++) {
            a[i] = x--;
        }

        for (int i : a)
            out.print(i + " ");
    }

    void reverseInt(int[] a) {
        int b;
        for (int i = 0, j = a.length - 1; i < j; i++, j--) {
            b = a[i];
            a[i] = a[j];
            a[j] = b;
        }
    }

    private void solveE() throws IOException {

    }

    private void solveF() throws IOException {

    }

    private void solveG() throws IOException {

    }

    private void solveH() throws IOException {

    }

    class FastScanner {
        StringTokenizer st;
        BufferedReader br;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        boolean hasNext() throws IOException {
            return br.ready() || (st != null && st.hasMoreTokens());
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

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

        boolean hasNextLine() throws IOException {
            return br.ready();
        }
    }

    private void run() throws IOException {
        in = new FastScanner(System.in); // new FastScanner(new FileInputStream(".in"));
        out = new PrintWriter(System.out); // new PrintWriter(new FileOutputStream(".out"));

        solve();

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

    public static void main(String[] args) throws IOException {
        new Main().run();
    }
}