#include <iostream>

using namespace std;

int main()
{
	int total = 0; // test
	
	int n;
	while ( cin >> n ) {
		int count;
		if (n>36) count = 0;
		else if (n<=9) count = ((n+1)*(n+2)*(n+3))/6;
		else if (n<=18) count = ((n+1)*(n+2)*(n+3))/6 - 4*((n-9)*(n-8)*(n-7))/6;
		// 220 - ((n-10+1)*(n-10+2)*(n-10+3))/6 + \sum_{k=10}^n { (k+1)(k+2)/2 - 3*(k-9)(k-8)/2 }
		// 220 - ((n-10+1)*(n-10+2)*(n-10+3))/6 + \sum_{k=10}^n { -k^2 + 27k - 107 }
		// 220 - ((n-10+1)*(n-10+2)*(n-10+3))/6 + \sum_{k=1}^n { -k^2 + 27k - 107 } - \sum_{k=1}^9 { -k^2 + 27k - 107 }
		// 220 - ((n-9)*(n-8)*(n-7))/6        - n(n+1)(2n+1)/6      + 27 * n(n+1)/2   - 107n    - 9*10*11/6 + 27*9*10/2 - 107*9
		// 220 - (n^3 - 24n^2 + 191n - 504)/6 - (2n^3 + 3n^2 + n)/6 + (27n^2 + 27n)/2 - 107n    - 165 + 1215 - 963
		// 307 - (3n^3 - 21n^2 + 192n - 504)/6                      + (81n^2 + 81n)/6 - 642n/6
		// 307 - (3n^3 - 102n^2 + 753n - 504)/6
		// 307 - (n^3  -  34n^2 + 251n - 168)/2
		// (-n^3 + 34n^2 - 251n + 782)/2 
		
		// 220  + \sum_{k=10}^n { (k+1)(k+2)/2 } - 4 * ((n-10+1)*(n-10+2)*(n-10+3))/6
		// \sum_{k=1}^n { (k+1)(k+2)/2 }         - 4 * ((n-10+1)*(n-10+2)*(n-10+3))/6
		else if (n<=27) count = ((37-n)*(38-n)*(39-n))/6 - 4*((27-n)*(28-n)*(29-n))/6;
		else count = ((37-n)*(38-n)*(39-n))/6;
		
		cout << count << endl;
		
		total += count; // test
	}
	
	cout << "total = " << total << endl; // test
	
	return 0;
}