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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.File;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Atharva Nagarkar
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        JoltyScanner in = new JoltyScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        CThePhoneNumber solver = new CThePhoneNumber();
        solver.solve(1, in, out);
        out.close();
    }

    static class CThePhoneNumber {
        public void solve(int testNumber, JoltyScanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] a = new int[n];
            int bestCycle = 1;
            int minCost = (int) 1e9;
            for (int i = 1; i <= n; ++i) {
                int lis = i;
                int lds = (n / i) + (n % i == 0 ? 0 : 1);
                int cur = lis + lds;
                if (cur < minCost) {
                    minCost = cur;
                    bestCycle = i;
                }
            }
            int val = n;
            for (int i = 0; i < n; i += bestCycle) {
                int sub = 0;
                for (int j = i, k = 1; j < Math.min(n, i + bestCycle); ++j, ++k, ++sub) {
                    a[j] = val - (bestCycle - k);
                }
                val -= sub;
            }
            int rem = n - n % bestCycle;
            for (int i = rem, k = 1; i < n; ++i, ++k) a[i] = ++val;
            for (int i = 0; i < n; ++i) out.print(a[i] + " ");
        }

    }

    static class JoltyScanner {
        public int BS = 1 << 16;
        public char NC = (char) 0;
        public byte[] buf = new byte[BS];
        public int bId = 0;
        public int size = 0;
        public char c = NC;
        public double num = 1;
        public BufferedInputStream in;

        public JoltyScanner(InputStream is) {
            in = new BufferedInputStream(is, BS);
        }

        public JoltyScanner(String s) throws FileNotFoundException {
            in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
        }

        public char nextChar() {
            while (bId == size) {
                try {
                    size = in.read(buf);
                } catch (Exception e) {
                    return NC;
                }
                if (size == -1) return NC;
                bId = 0;
            }
            return (char) buf[bId++];
        }

        public int nextInt() {
            return (int) nextLong();
        }

        public long nextLong() {
            num = 1;
            boolean neg = false;
            if (c == NC) c = nextChar();
            for (; (c < '0' || c > '9'); c = nextChar()) {
                if (c == '-') neg = true;
            }
            long res = 0;
            for (; c >= '0' && c <= '9'; c = nextChar()) {
                res = (res << 3) + (res << 1) + c - '0';
                num *= 10;
            }
            return neg ? -res : res;
        }

    }
}

