[codeforces] A - Clear Symmetry 规律

A - Clear Symmetry

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 201A

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 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.

Sample Input

Input

4

Output

3

Input

9

Output

5

Hint

The figure below shows the matrices that correspond to the samples:

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int main(){int n;cin >> n;if(n==1){cout << "1" <<endl; return 0; } if(n==2){cout << "3" <<endl; return 0; } if(n==3){cout << "5" <<endl; return 0; } if(n%4==0){n/=4;for(int i=1;i<=1000000;i++){n -= i;if(n<=0){cout << i*2+1 <<endl;return 0;}}}if(n%4==1){n/=4;for(int i=1;i<=1000000;i++){n -= i;if(n<=0){cout << i*2+1 <<endl;return 0;}}}if(n%4==2){for(int i=1;i<=1000000;i++){n -= i*4;if(n<=0){cout << i*2+1 <<endl;return 0;}}}if(n%4==3){n--;for(int i=1;i<=1000000;i++){n -= i*4;if(n<=0){cout << i*2+1 <<endl;return 0;}}}return 0;}