#include <bits/stdc++.h>
const int maxi = 2000000000;
const int maxq = 1000000000;
const double eps = 1e-10;
const double pi = 3.1415926535897932;
const double inf = 1e+18;
const int mo = 1000000007;
using namespace std;
int b[11111], ans[11111], n;
bool f[1111][1111];
int main() {
  cin >> n;
  if (n == 2) {
    puts("3");
    return 0;
  }
  for (int k = 1; k <= 2 * n; k++)
    if (((k)&1)) {
      for (int i = 0; i <= 5; i++) b[i] = 0;
      if (((k)&1))
        for (int i = 1; i <= k; i++)
          for (int j = 1; j <= k; j++)
            if (((i + j) & 1) == 0)
              f[i][j] = 1;
            else
              f[i][j] = 0;
      int kol = 0;
      for (int i = 1; i <= k; i++)
        for (int j = 1; j <= k; j++)
          if (f[i][j]) {
            int sum = 2;
            if (((k)&1) && i == k / 2 + 1) sum--;
            if (((k)&1) && j == k / 2 + 1) sum--;
            if (sum)
              b[sum * 2]++;
            else
              b[1]++;
            kol++;
          }
      b[4] /= 4;
      b[2] /= 2;
      for (int i = 0; i <= b[1]; i++)
        for (int j = 0; j <= b[2]; j++)
          for (int kk = 0; kk <= b[4]; kk++) {
            if (ans[kol - i - j * 2 - kk * 4] == 0)
              ans[kol - i - j * 2 - kk * 4] = k;
          }
    }
  cout << ans[n] << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
const int maxi = 2000000000;
const int maxq = 1000000000;
const double eps = 1e-10;
const double pi = 3.1415926535897932;
const double inf = 1e+18;
const int mo = 1000000007;
using namespace std;
int b[11111], ans[11111], n;
bool f[1111][1111];
int main() {
  cin >> n;
  if (n == 2) {
    puts("3");
    return 0;
  }
  for (int k = 1; k <= 2 * n; k++)
    if (((k)&1)) {
      for (int i = 0; i <= 5; i++) b[i] = 0;
      if (((k)&1))
        for (int i = 1; i <= k; i++)
          for (int j = 1; j <= k; j++)
            if (((i + j) & 1) == 0)
              f[i][j] = 1;
            else
              f[i][j] = 0;
      int kol = 0;
      for (int i = 1; i <= k; i++)
        for (int j = 1; j <= k; j++)
          if (f[i][j]) {
            int sum = 2;
            if (((k)&1) && i == k / 2 + 1) sum--;
            if (((k)&1) && j == k / 2 + 1) sum--;
            if (sum)
              b[sum * 2]++;
            else
              b[1]++;
            kol++;
          }
      b[4] /= 4;
      b[2] /= 2;
      for (int i = 0; i <= b[1]; i++)
        for (int j = 0; j <= b[2]; j++)
          for (int kk = 0; kk <= b[4]; kk++) {
            if (ans[kol - i - j * 2 - kk * 4] == 0)
              ans[kol - i - j * 2 - kk * 4] = k;
          }
    }
  cout << ans[n] << endl;
  return 0;
}

```