Clear Symmetry CodeForces

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 A_i, 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: A_i, j = A_{n - i + 1, j} and A_i, j = A_{i, 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.

Example

Input

4

Output

3

Input

9

Output

5

Note

The figure below shows the matrices that correspond to the samples

题解：①当n是偶数时，n-1所用到1的个数比n多。

　　　②当n是奇数时，最多能用(n*n+1)/2个1。

　　　③能用奇数构成所有的x。

①和③没证明出来，画图归纳的。。。。。

注意：x=1,n=1; x=2,n=3; x=3,n=5;

#include<bits/stdc++.h>
using namespace std;

int x;
int a[]={1,5,13,25,41,61,85,113};
int b[]={1,3,5,7,9,11,13,15};

int main()
{   cin>>x;
    int p=0;
    if(x==3){ cout<<"5"<<endl; return 0; }
    while(a[p]<x) p++;
    cout<<b[p]<<endl;
}