#include <bits/stdc++.h>

using namespace std;
string multiply(string num1, string num2);
string sumOfGroup(string k) {
    string s=multiply(k,k);
    return(multiply(k,s));
}
string multiply(string num1, string num2)
{
    int n1 = num1.size();
    int n2 = num2.size();
    if (n1 == 0 || n2 == 0)
       return "0";
 
    // will keep the result number in vector
    // in reverse order
    vector<int> result(n1 + n2, 0);
 
    // Below two indexes are used to find positions
    // in result. 
    int i_n1 = 0; 
    int i_n2 = 0; 
 
    // Go from right to left in num1
    for (int i=n1-1; i>=0; i--)
    {
        int carry = 0;
        int n1 = num1[i] - '0';
 
        // To shift position to left after every
        // multiplication of a digit in num2
        i_n2 = 0; 
         
        // Go from right to left in num2             
        for (int j=n2-1; j>=0; j--)
        {
            // Take current digit of second number
            int n2 = num2[j] - '0';
 
            // Multiply with current digit of first number
            // and add result to previously stored result
            // at current position. 
            int sum = n1*n2 + result[i_n1 + i_n2] + carry;
 
            // Carry for next iteration
            carry = sum/10;
 
            // Store result
            result[i_n1 + i_n2] = sum % 10;
 
            i_n2++;
        }
 
        // store carry in next cell
        if (carry > 0)
            result[i_n1 + i_n2] += carry;
 
        // To shift position to left after every
        // multiplication of a digit in num1.
        i_n1++;
    }
 
    // ignore '0's from the right
    int i = result.size() - 1;
    while (i>=0 && result[i] == 0)
       i--;
 
    // If all were '0's - means either both or
    // one of num1 or num2 were '0'
    if (i == -1)
       return "0";
 
    // generate the result string
    string s = "";
    while (i >= 0)
        s += std::to_string(result[i--]);
 
    return s;
}

int main() {
    string k;
    cin >> k;
    string answer = sumOfGroup(k);
    cout << answer << endl;
    return 0;
}