Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.

Let's call matrix A clear if no two cells containing ones have a common side.

Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.

Let's define the sharpness of matrix A as the number of ones in it.

Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.

Input

The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.

Output

Print a single number — the sought value of n.

Examples

Input

4


Output

3


Input

9


Output

5

Note

The figure below shows the matrices that correspond to the samples:

<image>
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class B {
	static int[] stuff=  {1, 3, 5, 3, 3, 5, 5, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15};
	public static void main(String[] args) {
		MScanner sc = new MScanner();
		PrintWriter out = new PrintWriter(System.out);
		int N = sc.nextInt()-1;
		out.println(stuff[N]);
		out.close();
		
	}

	

	static class MScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public MScanner() {
			stream = System.in;
			// stream = new FileInputStream(new File("dec.in"));

		}

		int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

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

		int[] nextInt(int N) {
			int[] ret = new int[N];
			for (int a = 0; a < N; a++)
				ret[a] = nextInt();
			return ret;
		}
		long nextLong() {
			return Long.parseLong(next());
		}

		long[] nextLong(int N) {
			long[] ret = new long[N];
			for (int a = 0; a < N; a++)
				ret[a] = nextLong();
			return ret;
		}

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

		double[] nextDouble(int N) {
			double[] ret = new double[N];
			for (int a = 0; a < N; a++)
				ret[a] = nextDouble();
			return ret;
		}

		String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		String[] next(int N) {
			String[] ret = new String[N];
			for (int a = 0; a < N; a++)
				ret[a] = next();
			return ret;
		}

		String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}

		String[] nextLine(int N) {
			String[] ret = new String[N];
			for (int a = 0; a < N; a++)
				ret[a] = nextLine();
			return ret;
		}

	}
}
