#include<bits/stdc++.h>
using namespace std;
#define ff              first
#define ss              second
#define int             long long
#define pb              push_back
#define mp              make_pair
#define pii              pair<int,int>
#define vi              vector<int>
#define mii             map<int,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define mod             1000000007
#define inf             1e18
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(x)            int x; cin>>x; while(x--)
#define pw(b,p)         pow(b,p) + 0.1
#define __              ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define rep(i,a,b) 		for(int i=a;i<b;i++)
#define repb(i,a,b) 	for(int i=a;i>=b;i--)
#define endl			"\n"
#define m(a)		    memset(a,0,sizeof(a))
#define all(x)                          begin(x), end(x)

void fastscan(int &x){
bool neg = false;register int c;
x = 0;c = getchar();
if(c=='-'){neg = true;c = getchar();}
for(; (c>47 && c<58); c=getchar()){x = x *10 + c - 48;}
if(neg){x *= -1;}}

void file(){
	#ifndef ONLINE_JUDGE
	freopen("in.txt" , "r" , stdin);
	freopen("out.txt" , "w" , stdout);
	#endif
}

int bs(int a[] , int s , int e, int x){
	if(s<=e){
		int m = (s+e)/2;
		if(a[m]==x)
			return m;
		else if(a[m] > x)
			return bs(a,s,m-1,x);
		else return bs(a,m+1,e,x);
	}
	return -1;
}

#define tr(x) cout<<x<<endl;
#define tr2(x,y) cout<<x<<" "<<y<<endl;
#define tr3(x,y,z) cout<<x<<" "<<y<<" "<<z<<endl;
#define tr4(w,x,y,z) cout<<w<<" "<<x<<" "<<y<<" "<<z<<endl;
#define tr5(v,w,x,y,z) cout<<v<<" "<<w<<" "<<x<<" "<<y<<" "<<z<<endl;
#define tr6(u,v,w,x,y,z) cout<<u<<" "<<v<<" "<<w<<" "<<x<<" "<<y<<" "<<z<<endl;

#define deb(x) cout<<#x<<"="<<x<<endl
#define deb2(x,y) cout<<#x<<"="<<x<<","<<#y<<"="<<y<<endl
#define deb3(x,y,z) cout<<#x<<"="<<x<<","<<#y<<"="<<y<<","<<#z<<"="<<z<<endl
#define deba(a) for(auto &x:a) cout<<x<<" ";

#define sd(x) cin >> x;
#define sd2(x,y) cin >> x >> y;
#define sd3(x,y,z) cin >> x >> y >> z;
#define sd4(w,x,y,z) cin >> w >> x >> y >> z;

bool is_pal(string s) {return equal(all(s), s.rbegin());}

int mypow(int a, int b){        //(logn) fast exponentiation
    int res = 1;            // call int x = mypow(2,5);
    while (b > 0) {
        if (b & 1)
            res = (res * a)%mod;
        a = (a * a)%mod;
        b >>= 1;
    }
    return (res%mod);
}

void solve1(){
	//659 D2 B..restore permutations by merger..
	int n; cin>>n;
	map<int,int>m;
	for(int i=0 ; i<n*2; ++i){
		int x; sd(x);
		m[x]++;
		if(m[x]==1){
			cout<<x <<" ";
		}
		else continue;
	}
	cout << endl;
}

void solve2(){
	// 659 D2 C => mention in notes...
	/*	

	PICKING ELEMENTS FROM THE TERMINALS OF AN ARRAY IF FORMS A SORTED SEQUENCE, THEN IT IS EITHER :
		1. INCREASING
		2. DECREASING
		3. FORMS A PEAK i.e (increasing ... max ... decreasing)
		4. HAD THE NEW SEQUENCE BEEN DECREASING, THEN WE WOULD LOOK FOR A "V" PATTERN IN THE ARRAY..i.e a ditch instead of a peak..

	Algo:
		remove the longest prefix such that the resulting array satisfies the following:
			1. appending terminal (either side) elements to a new array results in the new array being sorted always..
			2. observe that for this, the original array (excluding the prefix to be removed) must be of the follwing forms
				a. Increasing completely (we pick from the start and keep appending)
				b. completely Decreasing, (we pick from the end and keep appending)
				c. should have behaviour similar to a ** PEAK **	(very important observation !!)
						=> [a[0] <= a[1] <= ....a[mid].... >= a[i] >= a[i+1] ...a[end]]
			3. thus we move from right to left and while the above conditions hold, we keep going back,
				the moment this is violated, we remove from the start to the current index (prefix to be removed)..
	*/
	int n; cin>>n;
	int a[n]; rep(i,0,n) sd(a[i]);
	int pos=n-1;
	while(pos > 0 and a[pos-1] >= a[pos]) --pos;
	while(pos > 0 and a[pos-1] <= a[pos]) --pos;
	cout << pos << endl;
}

void solve3(){
	// 656 D3 A
	/*
		given (x,y,x), find (a,b,c) s.t max(a,b)=x ; max(b,c) = y; max(a,c) = z;
		if(x<= y <= z) => m(a,b) <= m(b,c) <= m(a,c) => (if(b>a) then (c>b)) or if(a>b) then (c>a))
		therefore, a value always repeats => take into a[] , sort(a,a+3).
		if(a[1] != a[2]) return -1;
		else cout << a[0] << a[0] << a[2];	
	*/
	int a[3]; rep(i,0,3) sd(a[i]);
	sort(a,a+3);
	if(a[1] != a[2]){
		cout << "NO" << endl; return;
	}
	cout << "YES"<<endl;
	tr3(a[0] , a[0] ,a[1]);
}

void solve4(){
	//neighbouring grid..
	int n,m; cin>>n>>m;
	int a[n][m];
	rep(i,0,n){
		rep(j,0,m){
			sd(a[i][j]);
		}
	}
	int f=0;
	for(int i=0 ; i<n ; ++i){
		for(int j=0; j<m ; ++j){
			int lvl=0;
			if(i>0) ++lvl;
			if(i<n-1) ++lvl;
			if(j>0) ++lvl;
			if(j<m-1) ++lvl;
			if(a[i][j] > lvl){
				f=1; break;
			}
		}
		if(f){
			break;
		}
	}
	if(f){
		cout <<"NO" <<endl;
	}
	else{
		cout << "YES" << endl;
		rep(i,0,n){
			rep(j,0,m){
				int lvl=0;
				if(i>0) ++lvl;
				if(i<n-1) ++lvl;
				if(j>0) ++lvl;
				if(j<m-1) ++lvl;
				a[i][j] = lvl;
				cout << a[i][j] << " ";
			}
			cout << endl;
		}
	}
}

void solve5(){
	//CF ED90 A
	int a,b,c; cin >> a >> b >> c;
	//check the cost of 1 donut...cuz in first..
	if(a<c) cout << 1 << " ";
	else cout << -1 << " ";

	//check the cost of b donuts, cuz in second..
	if(a*b>c) cout << b << " ";
	else cout << -1 <<" ";

	cout << endl;
}

void atcoderD(){
	// sequence is a[0] + a[1] + a[1] + a[2] + a[2] + a[3] + a[3]....
	int n; cin >> n;
	vi a(n+1); rep(i,0,n) sd(a[i]);
	int ans=0;
	sort(all(a));
	reverse(all(a));
	ans+=a[0];	
	for(int i=2 ; i<n ; ++i){
		ans+=(a[(int)floor(i/2)]);
	}
	tr(ans);
}

int32_t main(){	
	__;
	file();
	int n; sd(n);
	cout << n+n*n + n*n*n << endl;
	return 0;
}




	

Measured memory_bytes: 3656