/*<アイデア> dp[桁数][未満フラグ][数字和]
dp[i+1][j || d<D][k+d]+=dp[i][j][k]*/
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
typedef long long ll;
ll mod = 1000000009;
ll dp[2][300][2][2010];
void make_dp(const string &S, int a) {
	dp[a][0][0][0] = 1;
	for (int i = 0; i < S.length(); i++) {
		int D = S[i]-'0';
		for (int j = 0; j < 2; j++) {
			for (int k = 0; k < 2000; k++) {
				for (int d = 0; d <= (j ? 9 : D); d++) {
					dp[a][i + 1][j || (d < D)][k + d] += dp[a][i][j][k];
					dp[a][i + 1][j || (d < D)][k + d] %= mod;
				}
			}
		}
	}
}
int main() {
	ll ans = 0;
	string S, T;
	cin >> S >> T;
	memset(dp, 0, sizeof(dp));
	make_dp(S, 0);
	make_dp(T, 1);
	for (int k = 0; k < 2000; k++) {
		ll p = dp[0][S.length()][0][k] + dp[0][S.length()][1][k];
		ll q = dp[1][T.length()][0][k] + dp[1][T.length()][1][k];
		ans += p * q;
		ans %= mod;
	}
	cout << (ans + mod - 1) % mod << endl;
}