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#include <bits/stdc++.h>
using namespace std;
const int N = 10;
//数字0-9需要的火柴个数,这个字典妙啊~，成为解决火柴棍难题的关键~
int a[N] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
// 根据数字获取对应的火柴棍数量
int getNum(int n) {
if (n == 0) return a[0];
int s = 0;
while (n) {
int m = n % 10;
s += a[m];
n /= 10;
}
return s;
int n, cnt;
int main() {
//等号2个，加号2个
//给n根火柴棍，要求全部用上
//n的极限值是24，那么去掉4根就剩下20根，就是3个数加在一起需要20根
cin >> n;
for (int i = 0; i <= 999; i++)
for (int j = 0; j <= 999; j++)
if (getNum(i) + getNum(j) + getNum(i + j) == n - 4) cnt++;
cout << cnt << endl;
return 0;
