Description
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 nand 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.
Sample Input
4
3
9
5
Hint
The figure below shows the matrices that correspond to the samples:

#include<stdio.h>
int main()
{
int ans[18]={0},test;
for(int i=1;i<18;i+=2)
{
ans[i]=i*(i/2+1)-i/2;
}
while(scanf("%d",&test)!=EOF)
{
for(int i=1;i<101;i+=2)
{
if(test==3){printf("5\n");break;}
else
if(test<=ans[i])
{
printf("%d\n",i);
break;
}
}
}
return 0;
}