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>
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
string lesh[10];
string tp[20][30];
int n;
int XX;
int Solve(int ord[4], int c, int q) {
  string tmp = "";
  for (int i = 0; i < q; i++) tmp += tp[c][i];
  int cc = 0;
  int prev = 0;
  for (int i = 0; i < tmp.length(); i++) {
    int ci = ord[cc];
    while ((i - prev) < lesh[ci].length() && tmp[i] == lesh[ci][i - prev]) i++;
    if (i - prev == lesh[ci].length()) cc++;
    if (cc == n) break;
    prev = i;
  }
  if (cc != n) {
    return -1;
  }
  int x = 0;
  for (int i = 0; i < 4; i++) {
    if (ord[i] != i && ord[i] != -1) x++;
  }
  x *= n - 1;
  XX = x;
  int res = n * (n - 1) / 2 + 1 - x;
  return res;
}
int main() {
  scanf("%d", &n);
  int res = 0;
  if (n == 1) {
    printf("1");
    return 0;
  }
  if (n == 3) {
    printf("5");
    return 0;
  }
  if (n <= 5) {
    printf("3");
    return 0;
  }
  if (n < 14) {
    printf("5");
    return 0;
  }
  if (n < 26) {
    printf("7");
    return 0;
  }
  if (n < 42) {
    printf("9");
    return 0;
  }
  if (n < 62) {
    printf("11");
    return 0;
  }
  if (n < 86) {
    printf("13");
    return 0;
  }
  if (n < 101) {
    printf("15");
    return 0;
  }
  return 0;
}
