#include <bits/stdc++.h>
using namespace std;
void optimise() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long int mod = 1e9 + 7;
void normalize(long long int &a) { a = (a + mod) % mod; }
long long int modmul(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a * b) % mod;
}
long long int modadd(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a + b) % mod;
}
long long int modsub(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a - b + mod) % mod;
}
long long int me(long long int x, long long int n) {
  x %= mod;
  if (n == 0) return 1;
  long long int u = me(x, n / 2) % mod;
  u = (u * u) % mod;
  if (n % 2) u = (u * x) % mod;
  return u;
}
long long int me1(long long int x, long long int n) {
  if (n == 0) return 1;
  long long int u = me1(x, n / 2);
  u = u * u;
  if (n % 2) u = u * x;
  return u;
}
inline long long int fact(long long int r) {
  long long int ans = 1;
  while (r) ans = modmul(ans, r), r--;
  return ans;
}
inline long long int modInv(long long int a) { return me(a, mod - 2); }
inline long long int modDiv(long long int a, long long int b) {
  return modmul(a, modInv(b));
}
long long int __gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return __gcd(b, a % b);
}
void solve() {
  long long int n;
  cin >> n;
  if (n == 1) {
    cout << 1;
    return;
  }
  if (n == 3) {
    cout << 5;
    return;
  }
  if (n <= 4) {
    cout << 3;
    return;
  }
  if (n == 5) {
    cout << 3;
    return;
  }
  long long int i = 5;
  while (1) {
    long long int k1 = (i * i + 1) / 2;
    if (n <= k1) {
      cout << i;
      return;
    }
    i += 2;
  }
}
signed main() {
  optimise();
  long long int t;
  t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}

### Prompt

Please formulate 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;
void optimise() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long int mod = 1e9 + 7;
void normalize(long long int &a) { a = (a + mod) % mod; }
long long int modmul(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a * b) % mod;
}
long long int modadd(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a + b) % mod;
}
long long int modsub(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a - b + mod) % mod;
}
long long int me(long long int x, long long int n) {
  x %= mod;
  if (n == 0) return 1;
  long long int u = me(x, n / 2) % mod;
  u = (u * u) % mod;
  if (n % 2) u = (u * x) % mod;
  return u;
}
long long int me1(long long int x, long long int n) {
  if (n == 0) return 1;
  long long int u = me1(x, n / 2);
  u = u * u;
  if (n % 2) u = u * x;
  return u;
}
inline long long int fact(long long int r) {
  long long int ans = 1;
  while (r) ans = modmul(ans, r), r--;
  return ans;
}
inline long long int modInv(long long int a) { return me(a, mod - 2); }
inline long long int modDiv(long long int a, long long int b) {
  return modmul(a, modInv(b));
}
long long int __gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return __gcd(b, a % b);
}
void solve() {
  long long int n;
  cin >> n;
  if (n == 1) {
    cout << 1;
    return;
  }
  if (n == 3) {
    cout << 5;
    return;
  }
  if (n <= 4) {
    cout << 3;
    return;
  }
  if (n == 5) {
    cout << 3;
    return;
  }
  long long int i = 5;
  while (1) {
    long long int k1 = (i * i + 1) / 2;
    if (n <= k1) {
      cout << i;
      return;
    }
    i += 2;
  }
}
signed main() {
  optimise();
  long long int t;
  t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}

```