#include <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
typedef long long ll;
const ll mod = 1000000009;
ll dp1[202][2][2000]; // 桁数、小さいことが確定しているか、各桁の和
ll dp2[202][2][2000];
string M, D;
void calc1(){
  int len = M.length();
  FOR(i,0,len+1) FOR(j,0,2) FOR(k,0,2000) {
    dp1[i][j][k] = 0;
  }
  dp1[0][0][0] = 1;
  FOR(i,0,len) FOR(j,0,2) FOR(k,0,2000) {
    int lim = j ? 9 : M[i] - '0';
    FOR(d,0,lim+1) {
      if(k+d>=2000) continue;
      (dp1[i+1][j||d<lim][k+d] += dp1[i][j][k]) %= mod;
    }
  }
}
void calc2(){
  int len = D.length();
  FOR(i,0,len+1) FOR(j,0,2) FOR(k,0,2000) {
    dp2[i][j][k] = 0;
  }
  dp2[0][0][0] = 1;
  FOR(i,0,len) FOR(j,0,2) FOR(k,0,2000) {
    int lim = j ? 9 : D[i] - '0';
    FOR(d,0,lim+1) {
      if(k+d>=2000) continue;
      (dp2[i+1][j||d<lim][k+d] += dp2[i][j][k]) %= mod;
    }
  }
}
int main()
{
  cin >> M >> D;
  calc1();
  calc2();
  ll ans = 0;
  FOR(k,1,2000) {
    ans += ((dp1[M.length()][0][k]+dp1[M.length()][1][k])%mod) * ((dp2[D.length()][0][k]+dp2[D.length()][1][k])%mod);
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}