#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:255000000")
bool firstout = 1;
template <class T>
T &minn(T &a, T b) {
  if (b < a) a = b;
  return a;
}
template <class T>
T &maxx(T &a, T b) {
  if (a < b) a = b;
  return a;
}
int &madd(int &a, int b) {
  a += b;
  if (a >= 1000000009) a -= 1000000009;
  return a;
}
int &msub(int &a, int b) {
  a -= b;
  if (a < 0) a += 1000000009;
  return a;
}
int &mmult(int &a, int b) { return a = (long long)a * b % 1000000009; }
int mdiv(long long a, long long b, long long m) {
  a = (a % m + m) % m;
  b = (b % m + m) % m;
  if (a % b == 0) return a / b;
  return (a + m * mdiv(-a, m, b)) / b;
}
int n, m;
int A[1012];
int main() {
  int i, j, k;
  char c;
  int a, d;
  int ts;
  for (ts = 1; scanf("%"
                     "d",
                     &(n)) > 0;
       ++ts) {
    if (n == 3)
      printf(
          "%"
          "d",
          (5)),
          printf("\n"), firstout = 1;
    else {
      for (i = (1); i < (1012345678); i += (2))
        if ((((i) * (i)) + 1) / 2 >= n) break;
      printf(
          "%"
          "d",
          (i));
      printf("\n"), firstout = 1;
    }
    int res = 0;
  }
  return 0;
}

### Prompt

Generate a cpp solution to 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>
using namespace std;
#pragma comment(linker, "/STACK:255000000")
bool firstout = 1;
template <class T>
T &minn(T &a, T b) {
  if (b < a) a = b;
  return a;
}
template <class T>
T &maxx(T &a, T b) {
  if (a < b) a = b;
  return a;
}
int &madd(int &a, int b) {
  a += b;
  if (a >= 1000000009) a -= 1000000009;
  return a;
}
int &msub(int &a, int b) {
  a -= b;
  if (a < 0) a += 1000000009;
  return a;
}
int &mmult(int &a, int b) { return a = (long long)a * b % 1000000009; }
int mdiv(long long a, long long b, long long m) {
  a = (a % m + m) % m;
  b = (b % m + m) % m;
  if (a % b == 0) return a / b;
  return (a + m * mdiv(-a, m, b)) / b;
}
int n, m;
int A[1012];
int main() {
  int i, j, k;
  char c;
  int a, d;
  int ts;
  for (ts = 1; scanf("%"
                     "d",
                     &(n)) > 0;
       ++ts) {
    if (n == 3)
      printf(
          "%"
          "d",
          (5)),
          printf("\n"), firstout = 1;
    else {
      for (i = (1); i < (1012345678); i += (2))
        if ((((i) * (i)) + 1) / 2 >= n) break;
      printf(
          "%"
          "d",
          (i));
      printf("\n"), firstout = 1;
    }
    int res = 0;
  }
  return 0;
}

```