#include #include #include #include using namespace std; using namespace __gnu_pbds; template using ordered_set = tree, rb_tree_tag, tree_order_statistics_node_update>; struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<=0;i--) #define FOR(i,a,b) for(ll i=a;i<=ll(b);i++) #define FORD(i,a,b) for(ll i=a;i>=ll(b);i--) #define overload4(_1,_2,_3,_4,name,...) name #define overload3(_1,_2,_3,name,...) name #define rep1(n) for(ll i=0;i(a);) #define rrep4(i,a,b,c) for(ll i=(a)+((b)-(a)-1)/(c)*(c);i>=(a);i-=c) #define rrep(...) overload4(__VA_ARGS__,rrep4,rrep3,rrep2,rrep1)(__VA_ARGS__) #define ALL(v) v.begin(), v.end() #define endl "\n" #define fi first #define se second #define popcount(bit) __builtin_popcount(bit) #define popcountll(bit) __builtin_popcountll(bit) #define pb push_back #define eb emplace_back #define sz(x) ((ll)(x).size()) using namespace atcoder; using P = pair; using PL = pair; using Graph = vector>; typedef long long ll; template inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; } template inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; } inline void Yes(bool b = true) { cout << (b ? "Yes" : "No") << '\n'; } inline void YES(bool b = true) { cout << (b ? "YES" : "NO") << '\n'; } inline void OKNG(bool b = true) { cout << (b ? "OK" : "NG") << '\n'; } const int dx[4] = {1, 0, -1, 0}; const int dy[4] = {0, 1, 0, -1}; const int fx[8] = {0, 1, 1, 1, 0, -1, -1, -1}; const int fy[8] = {1, 1, 0, -1, -1, -1, 0, 1}; template const auto INF = numeric_limits::max()/2; using mint = static_modint<1000000009>; mint dp1[201][1801][2]; mint dp2[201][1801][2]; int main() { string M, D; cin >> M >> D; int n = M.size(); int m = D.size(); dp1[0][0][0] = 1; dp2[0][0][0] = 1; rep(i, n)rep(j, 1801)rep(k, 2){ if(dp1[i][j][k] == 0)continue; int nd = M[i] - '0'; rep(d, 10){ int nk = k; if(j + d > 1800)continue; if(k == 0){ if(d < nd)nk = 1; if(d > nd)continue; } dp1[i+1][j + d][nk] += dp1[i][j][k]; } } rep(i, m)rep(j, 1801)rep(k, 2){ if(dp2[i][j][k] == 0)continue; int nd = D[i] - '0'; rep(d, 10){ int nk = k; if(k == 0){ if(d < nd)nk = 1; if(d > nd)continue; } dp2[i+1][j + d][nk] += dp2[i][j][k]; } } mint ans = 0; rep3(i, 1, 1801){ ans += (dp1[n][i][0] + dp1[n][i][1]) * (dp2[m][i][0] + dp2[m][i][1]); } cout << ans.val() << endl; }