Problem:
Score : 100 points

Problem Statement

It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.

Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Constraints

1 \leq r \leq 100

r is an integer.

Input

Input is given from Standard Input in the following format:

r

Output

Print an integer representing the area of the regular dodecagon.

Sample Input 1

4

Sample Output 1

48

The area of the regular dodecagon is 3 \times 4^2 = 48.

Sample Input 2

15

Sample Output 2

675

Sample Input 3

80

Sample Output 3

19200

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
** Vishal Raj Roy
** Indian Institute of Technology Kharagpur
*/

#include <bits/stdc++.h>

using namespace std;

#define posLSB(X) __builtin_ctz(X)
#define num1bit(X) __builtin_popcount(X)
#define numlead0(X) __builtin_clz(X)
#define gcd(X,Y) __gcd(X,Y)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define ifpresent(CON,VAL) (CON.find(VAL) != CON.end())
#define all(x) x.begin(),x.end()
#define at(X,N) get<N>(X)
#define ins insert
#define xx first
#define yy second
#define sz(x) ((int)x.size())
#define count_ones __builtin_popcountl
#define nl cout<<"\n";
#define name(X) (#X)
#define watch(X) cout << (#X) << " is " << (X) << "\n"
#define sqr(x) ((x)*(x))
#define mod(x, m) ((((x) % (m)) + (m)) % (m))
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
#define mem(x,val) memset((x),(val),sizeof(x))
#define rite(X) freopen(X,"w",stdout)
#define read(X) freopen(X,"r",stdin)
//FOR(i,1,10) gives 1,2,3,....,9 and FOR(i,10,1) gives 9,8,....,1 also FOR(it, end(v), begin(v)) and FOR(it, begin(v), end(v))
#define FOR(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define SEL(X,C) for( auto & X : C )
#define ima INT_MAX
#define imi INT_MIN
#define lma LLONG_MAX
#define lmi LLONG_MIN
using st = string; typedef long long int li; typedef pair<int, int> ii; typedef pair<int, ii> iii; typedef tuple<int,int> i2; typedef tuple<int,int,int> i3; typedef vector<int> vi; typedef vector<ii> vii; typedef vector<i2> vi2; typedef vector<i3> vi3; typedef vector<iii> viii; typedef vector<vi> vvi; typedef vector<vvi> vvvi; typedef pair<li, li> ll; typedef pair<li, ll> lll; typedef tuple<li,li> l2; typedef tuple<li,li,li> l3; typedef vector<li> vl; typedef vector<ll> vll; typedef vector<l2> vl2; typedef vector<l3> vl3; typedef vector<lll> vlll; typedef vector<vl> vvl; typedef vector<vvl> vvvl; typedef double db; typedef complex<db> cd; typedef vector<cd> vcd; template <class S> using ve = vector<S>; template <class S> using gr = greater<S>; template <class S> using le = less<S>;

const double pi = 2 * acos(0.0);
const int oo = 0x3f3f3f3f; // don't use for long long
const double inf = 1.0/0.0;


int main() {
    ios_base::sync_with_stdio(0);
    int r; cin>>r; cout<<3*r*r<<"\n";
    return 0;
}