# include # include # include # include # include # include # include # include # include # include # include # include # include # include using namespace std; using LL = long long; using ULL = unsigned long long; constexpr int INF = 2000000000; constexpr int HINF = INF / 2; constexpr double DINF = 100000000000000000.0; constexpr long long LINF = 9223372036854775807; constexpr long long HLINF = 4500000000000000000; const double PI = acos(-1); int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 }; #define FOR(i,a,b) for(int i=(a);i<(b);++i) #define REP(i,n) FOR(i,0,n) #define ALL(x) (x).begin(),(x).end() #define UNIQ(c) (c).erase(unique(ALL((c))), end((c))) #define mp make_pair #define eb emplace_back //typedef pair P; //typedef pair PP; LL dp[101010][2][2][3][8];//keta,less,has3,mod3,mod8 LL dp2[101010][2][2][3][8]; string a; int n; const LL MOD = 1e9 + 7; int main() { cin >> a; for (int c = a.size() - 1; c >= 0; c--) { if (a[c] != '0') { a[c]--; break; } else { a[c] = '9'; } } n = a.length(); dp[0][0][0][0][0] = 1; REP(i, n)REP(j, 2)REP(k, 2)REP(l, 3)REP(m, 8) { int lim = (j ? 9 : a[i] - '0'); REP(d, lim + 1) { (dp[i + 1][j || d < lim][k || d == 3][(l + d) % 3][(10 * m + d) % 8] += dp[i][j][k][l][m]) %= MOD; } } LL ans = 0; REP(j, 2)REP(k, 2)REP(l, 3)REP(m, 8)if ((k || l == 0) && m != 0)(ans += dp[n][j][k][l][m]) %= MOD; string b; cin >> b; n = b.length(); dp2[0][0][0][0][0] = 1; REP(i, n)REP(j, 2)REP(k, 2)REP(l, 3)REP(m, 8) { int lim = (j ? 9 : b[i] - '0'); REP(d, lim + 1) { (dp2[i + 1][j || d < lim][k || d == 3][(l + d) % 3][(10 * m + d) % 8] += dp2[i][j][k][l][m]) %= MOD; } } LL ans2 = 0; REP(j, 2)REP(k, 2)REP(l, 3)REP(m, 8)if ((k || l == 0) && m != 0)(ans2 += dp2[n][j][k][l][m]) %= MOD; cout << (ans2 - ans + MOD) % MOD << endl; //system("pause"); }