#include <bits/stdc++.h>

using namespace std;

// 棋盘覆盖问题原理
// https://blog.csdn.net/m0_51005809/article/details/120112637

const int N = 1e5 + 10;
int a[N], al;

void mul(int a[], int &al, int b) {
    int t = 0;
    for (int i = 1; i <= al; i++) {
        t += a[i] * b;
        a[i] = t % 10;
        t /= 10;
    }
    while (t) {
        a[++al] = t % 10;
        t /= 10;
    }
    while (al > 1 && !a[al]) al--;
}

int r;
void div(int a[], int &al, int b, int &r) {
    r = 0;
    for (int i = al; i >= 1; i--) {
        r = r * 10 + a[i];
        a[i] = r / b;
        r %= b;
    }
    while (al > 1 && !a[al]) al--;
}

int main() {
    int k;
    cin >> k;
    //通过高精度计算4^k
    a[1] = 1, al = 1;
    for (int i = 1; i <= k; i++) mul(a, al, 4);

    div(a, al, 3, r);
    for (int i = al; i; i--) printf("%d", a[i]);
    return 0;
}