#include <bits/stdc++.h>
using namespace std;
int testCase() {
  int x;
  if (scanf("%d", &x) != 1) {
    return -1;
  }
  if (x == 1) {
    return 1;
  }
  if (x == 2) {
    return 3;
  }
  for (int n = 3; n < 10000; ++n) {
    if (n % 2 == 0) {
      int nn = n / 2 - 1;
      int k = (nn * nn + 1) / 2;
      if (x % 4 == 0 && x / 4 == k) {
        return n;
      }
    } else {
      int nn = n / 2;
      if (x % 2 == 0) {
        if (nn % 2 == 0) {
          int xx = x;
          int k = (nn * nn) / 2;
          xx -= min(k, xx / 4) * 4;
          int l = (nn / 2) * 2;
          xx -= min(l, xx / 2) * 2;
          if (xx == 0) {
            return n;
          }
        } else {
          int xx = x;
          int k = (nn * nn + 1) / 2;
          xx -= min(k, xx / 4) * 4;
          int l = (nn / 2) * 2;
          xx -= min(l, xx / 2) * 2;
          if (xx == 0) {
            return n;
          }
        }
      }
      {
        int xx = x;
        if (x % 2 != 0) {
          xx--;
        }
        int k = (nn * nn + 1) / 2;
        xx -= min(k, xx / 4) * 4;
        int l = (nn / 2) * 2;
        xx -= min(l, xx / 2) * 2;
        if (xx == 0) {
          return n;
        }
      }
    }
  }
  assert(false);
  return -1;
}
int main() {
  for (int ret; (ret = testCase()) != -1;) {
    printf("%d\n", ret);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
int testCase() {
  int x;
  if (scanf("%d", &x) != 1) {
    return -1;
  }
  if (x == 1) {
    return 1;
  }
  if (x == 2) {
    return 3;
  }
  for (int n = 3; n < 10000; ++n) {
    if (n % 2 == 0) {
      int nn = n / 2 - 1;
      int k = (nn * nn + 1) / 2;
      if (x % 4 == 0 && x / 4 == k) {
        return n;
      }
    } else {
      int nn = n / 2;
      if (x % 2 == 0) {
        if (nn % 2 == 0) {
          int xx = x;
          int k = (nn * nn) / 2;
          xx -= min(k, xx / 4) * 4;
          int l = (nn / 2) * 2;
          xx -= min(l, xx / 2) * 2;
          if (xx == 0) {
            return n;
          }
        } else {
          int xx = x;
          int k = (nn * nn + 1) / 2;
          xx -= min(k, xx / 4) * 4;
          int l = (nn / 2) * 2;
          xx -= min(l, xx / 2) * 2;
          if (xx == 0) {
            return n;
          }
        }
      }
      {
        int xx = x;
        if (x % 2 != 0) {
          xx--;
        }
        int k = (nn * nn + 1) / 2;
        xx -= min(k, xx / 4) * 4;
        int l = (nn / 2) * 2;
        xx -= min(l, xx / 2) * 2;
        if (xx == 0) {
          return n;
        }
      }
    }
  }
  assert(false);
  return -1;
}
int main() {
  for (int ret; (ret = testCase()) != -1;) {
    printf("%d\n", ret);
  }
  return 0;
}

```