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 arr[105][105];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int x;
  cin >> x;
  for (int n = 1; n <= 100; n++) {
    int tmp = x;
    memset(arr, 0, sizeof arr);
    for (int row = 1; row < (n + 1) / 2; row++) {
      int st;
      if (row & 1)
        st = 1;
      else
        st = 2;
      for (int col = st; col < (n + 1) / 2; col += 2) {
        if (x >= 4) {
          arr[row][col] = 1, arr[n - row + 1][col] = 1,
          arr[row][n - col + 1] = 1, arr[n - row + 1][n - col + 1] = 1;
          x -= 4;
        }
      }
    }
    if (n & 1) {
      int cur_row = (n + 1) / 2;
      int st;
      if (cur_row & 1)
        st = 1;
      else
        st = 2;
      for (int col = st; col < (n + 1) / 2; col += 2) {
        if (x >= 2)
          arr[cur_row][col] = 1, arr[cur_row][n - col + 1] = 1, x -= 2;
      }
      int cur_col = cur_row;
      if (cur_col & 1)
        st = 1;
      else
        st = 2;
      for (int row = st; row < (n + 1) / 2; row += 2) {
        if (x >= 2) arr[row][cur_col] = arr[n - row + 1][cur_col] = 1, x -= 2;
      }
      if (x > 0) arr[cur_row][cur_col] = 1, x--;
    }
    if (x == 0) {
      cout << n;
      return 0;
    }
    x = tmp;
    memset(arr, 0, sizeof arr);
    for (int row = 1; row < (n + 1) / 2; row++) {
      int st;
      if (row & 1)
        st = 2;
      else
        st = 1;
      for (int col = st; col < (n + 1) / 2; col += 2) {
        if (x >= 4) {
          arr[row][col] = 1, arr[n - row + 1][col] = 1,
          arr[row][n - col + 1] = 1, arr[n - row + 1][n - col + 1] = 1;
          x -= 4;
        }
      }
    }
    if (n & 1) {
      int cur_row = (n + 1) / 2;
      int st;
      if (cur_row & 1)
        st = 2;
      else
        st = 1;
      for (int col = st; col < (n + 1) / 2; col += 2) {
        if (x >= 2)
          arr[cur_row][col] = 1, arr[cur_row][n - col + 1] = 1, x -= 2;
      }
      int cur_col = cur_row;
      if (cur_col & 1)
        st = 2;
      else
        st = 1;
      for (int row = st; row < (n + 1) / 2; row += 2) {
        if (x >= 2) arr[row][cur_col] = arr[n - row + 1][cur_col] = 1, x -= 2;
      }
    }
    if (x == 0) {
      cout << n;
      return 0;
    }
    x = tmp;
  }
  return 0;
}
