#include <stdio.h>

// Program to count how many 8 character combinations are necessary for brute force

// Constants to remove magic numbers
#define Z 25

int main (void) {
    static const short int max[26] = {4, 3, 3, 4, 4, 4, 4, 3, 3, 2, 3, 4, 3, 4, 4, 3, 1, 4, 5, 3, 4, 2, 3, 2, 3, 4}; // from STDERR of filter.pl
    short int have[26] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
    int combo = 0;
    int i, j, k, l, m, n, o, p;

    for (i = 0; i <= Z; i++) {
        if (++have[i] > max[i]) {
            --have[i];
            continue;
        }
        for (j = i; j <= Z; j++) {
            if (++have[j] > max[j]) {
                --have[j];
                continue;
            }
            for (k = j; k <= Z; k++) {
                if (++have[k] > max[k]) {
                    --have[k];
                    continue;
                }
                for (l = k; l <= Z; l++) {
                    if (++have[l] > max[l]) {
                        --have[l];
                        continue;
                    }
                    for (m = l; m <= Z; m++) {
                        if (++have[m] > max[m]) {
                            --have[m];
                            continue;
                        }
                        for (n = m; n <= Z; n++) {
                            if (++have[n] > max[n]) {
                                --have[n];
                                continue;
                            }
                            for (o = n; o <= Z; o++) {
                                if (++have[o] > max[o]) {
                                    --have[o];
                                    continue;
                                }
                                for (p = o; p <= Z; p++) {
                                    if (++have[p] > max[p]) {
                                        --have[p];
                                        continue;
                                    }
                                    ++combo;
                                    --have[p];
                                }
                                --have[o];
                            }
                            --have[n];
                        }
                        --have[m];
                    }
                    --have[l];
                }
                --have[k];
            }
            --have[j];
        }
        --have[i];
    }
    printf("%i\n", combo);
    return 0;
}