#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long p = 1, data = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    data = data * 10 + (ch ^ 48);
    ch = getchar();
  }
  return p * data;
}
inline long long qpow(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = a * r;
    a = a * a;
    b >>= 1;
  }
  return r;
}
inline long long gcd(long long a, long long b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b)
    x = 1, y = 0;
  else {
    exgcd(b, a % b, y, x);
    y -= x * (a / b);
  }
}
const int mod = 1e9 + 7;
int a[20];
int main() {
  int x = read();
  if (x == 3) {
    puts("5");
    return 0;
  }
  a[1] = 1, a[2] = 2;
  for (int i = 3; i <= 15; i += 2)
    a[i] = (i + 1) / 2 * 4 - 4 * (i & 1) + a[i - 2];
  for (int i = 1; i <= 15; i += 2) {
    if (a[i] >= x) {
      printf("%d\n", i);
      return 0;
    }
  }
  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>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long p = 1, data = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    data = data * 10 + (ch ^ 48);
    ch = getchar();
  }
  return p * data;
}
inline long long qpow(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = a * r;
    a = a * a;
    b >>= 1;
  }
  return r;
}
inline long long gcd(long long a, long long b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b)
    x = 1, y = 0;
  else {
    exgcd(b, a % b, y, x);
    y -= x * (a / b);
  }
}
const int mod = 1e9 + 7;
int a[20];
int main() {
  int x = read();
  if (x == 3) {
    puts("5");
    return 0;
  }
  a[1] = 1, a[2] = 2;
  for (int i = 3; i <= 15; i += 2)
    a[i] = (i + 1) / 2 * 4 - 4 * (i & 1) + a[i - 2];
  for (int i = 1; i <= 15; i += 2) {
    if (a[i] >= x) {
      printf("%d\n", i);
      return 0;
    }
  }
  return 0;
}

```