#include <bits/stdc++.h>
long long int ciel(double a) {
  long long int ret = a;
  if (a - ret > 0.0000000001) ret++;
  return ret;
}
long long int gcd(long long int a, long long int b) {
  if (a < b) return gcd(b, a);
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
double fgcd(double a, double b) {
  if (fabs(a) < fabs(b)) return fgcd(b, a);
  if (fabs(b) < 0.0001) return a;
  return fgcd(b, fmod(a, b));
}
bool db_db_cmp(double a, double b) {
  return (a + 0.0000000001 > b && a - 0.0000000001 < b);
}
long long int fact[100005];
void generate_factorials() {
  fact[0] = fact[1] = 1;
  for (int i = 2; i <= 100005 - 1; i++)
    fact[i] = ((fact[i - 1] * i) % 1000000007);
}
long long int pow(long long int n, long long int p) {
  if (p == 0) return 1;
  if (n <= 1) return n;
  long long int res = 1;
  while (p) {
    if (p & 1) res = res * n;
    n = n * n;
    p /= 2;
  }
  return res;
}
long long int mod_pow(long long int n, long long int p) {
  if (p == 0) return 1;
  if (n <= 1) return n;
  long long int res = 1;
  while (p) {
    if (p & 1) res = ((res * n) % 1000000007);
    n = ((n * n) % 1000000007);
    p /= 2;
  }
  return res;
}
long long int get_inverse(long long int n) {
  return mod_pow(n, 1000000007 - 2);
}
long long int get_nCr(long long int n, long long int r) {
  return ((fact[n] * get_inverse(((fact[r] * fact[n - r]) % 1000000007))) %
          1000000007);
}
using namespace std;
void preprocess() {}
void init() {}
void solve(int test_case) {
  init();
  int a;
  cin >> a;
  if (a == 1) {
    cout << 1 << endl;
    return;
  }
  if (a == 2) {
    cout << 3 << endl;
    return;
  }
  if (a == 3) {
    cout << 5 << endl;
    return;
  }
  int maxi = 1, n = 1;
  while (maxi < a) {
    n += 2;
    maxi = n / 2;
    maxi *= n;
    maxi = (maxi + 1) / 2;
    maxi *= 2;
    int temp = n / 2;
    maxi += n / 2;
    if (!(temp & 1)) {
      maxi++;
    }
  }
  cout << n << endl;
}
int main() {
  preprocess();
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
long long int ciel(double a) {
  long long int ret = a;
  if (a - ret > 0.0000000001) ret++;
  return ret;
}
long long int gcd(long long int a, long long int b) {
  if (a < b) return gcd(b, a);
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
double fgcd(double a, double b) {
  if (fabs(a) < fabs(b)) return fgcd(b, a);
  if (fabs(b) < 0.0001) return a;
  return fgcd(b, fmod(a, b));
}
bool db_db_cmp(double a, double b) {
  return (a + 0.0000000001 > b && a - 0.0000000001 < b);
}
long long int fact[100005];
void generate_factorials() {
  fact[0] = fact[1] = 1;
  for (int i = 2; i <= 100005 - 1; i++)
    fact[i] = ((fact[i - 1] * i) % 1000000007);
}
long long int pow(long long int n, long long int p) {
  if (p == 0) return 1;
  if (n <= 1) return n;
  long long int res = 1;
  while (p) {
    if (p & 1) res = res * n;
    n = n * n;
    p /= 2;
  }
  return res;
}
long long int mod_pow(long long int n, long long int p) {
  if (p == 0) return 1;
  if (n <= 1) return n;
  long long int res = 1;
  while (p) {
    if (p & 1) res = ((res * n) % 1000000007);
    n = ((n * n) % 1000000007);
    p /= 2;
  }
  return res;
}
long long int get_inverse(long long int n) {
  return mod_pow(n, 1000000007 - 2);
}
long long int get_nCr(long long int n, long long int r) {
  return ((fact[n] * get_inverse(((fact[r] * fact[n - r]) % 1000000007))) %
          1000000007);
}
using namespace std;
void preprocess() {}
void init() {}
void solve(int test_case) {
  init();
  int a;
  cin >> a;
  if (a == 1) {
    cout << 1 << endl;
    return;
  }
  if (a == 2) {
    cout << 3 << endl;
    return;
  }
  if (a == 3) {
    cout << 5 << endl;
    return;
  }
  int maxi = 1, n = 1;
  while (maxi < a) {
    n += 2;
    maxi = n / 2;
    maxi *= n;
    maxi = (maxi + 1) / 2;
    maxi *= 2;
    int temp = n / 2;
    maxi += n / 2;
    if (!(temp & 1)) {
      maxi++;
    }
  }
  cout << n << endl;
}
int main() {
  preprocess();
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}

```