// #include "atcoder/convolution"
#include "atcoder/dsu"
#include "atcoder/fenwicktree"
#include "atcoder/lazysegtree"
#include "atcoder/math"
#include "atcoder/maxflow"
#include "atcoder/mincostflow"
#include "atcoder/modint"
#include "atcoder/scc"
#include "atcoder/segtree"
#include "atcoder/string"
#include "atcoder/twosat"
using namespace atcoder;
#include <bits/stdc++.h>
using namespace std;
// #include <boost/multiprecision/cpp_int.hpp>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i,a,b) for (int i = (int)(a); i < (int)(b); i++)
#define all(v) v.begin(),v.end()
#define inc(x,l,r) ((l)<=(x)&&(x)<(r)) 
#define Unique(x) sort(all(x)), x.erase(unique(all(x)), x.end())
#define pcnt __builtin_popcountll
typedef long long ll;
#define int ll
using ld = long double;
using vi = vector<int>;
using vs = vector<string>;
using P = pair<int,int>;
using vp = vector<P>;
// using Bint = boost::multiprecision::cpp_int;
template<typename T1,typename T2> bool chmax(T1 &a, const T2 b) {if (a < b) {a = b; return true;} else return false; }
template<typename T1,typename T2> bool chmin(T1 &a, const T2 b) {if (a > b) {a = b; return true;} else return false; }
template<typename T> using priority_queue_greater = priority_queue<T, vector<T>, greater<T>>;
template<typename T> istream &operator>>(istream& is,vector<T> &v){for(T &in:v)is>>in;return is;}
template<typename T1,typename T2> ostream &operator<< (ostream &os, const pair<T1,T2> &p){os << p.first <<" "<<p.second;return os;}
ostream &operator<< (ostream &os, const modint1000000007 &m){os << m.val();return os;}
istream &operator>> (istream &is, modint1000000007 &m){ll in;is>>in;m=in;return is;}
ostream &operator<< (ostream &os, const modint998244353 &m){os << m.val();return os;}
istream &operator>> (istream &is, modint998244353 &m){ll in;is>>in;m=in;return is;}
template<class... T> void input(T&... a){(cin>> ... >> a);}
#ifdef LOCAL
template<typename T> ostream &operator<<(ostream &os,const vector<T> &v){os<<"\x1b[32m";rep(i,v.size())os<<v[i]<<(i+1!=v.size()?" ":"");os<<"\x1b[0m";return os;}
template<class T> void print(T& a){cout << "\x1b[32m"<< a<< '\n' << "\x1b[0m";}
template<class T,class... Ts> void print(const T&a, const Ts&... b){cout << "\x1b[32m" << a;(cout<<...<<(cout<<' ',b));cout<<'\n' << "\x1b[0m";}
#else
template<typename T> ostream &operator<<(ostream &os,const vector<T> &v){rep(i,v.size())os<<v[i]<<(i+1!=v.size()?" ":"");return os;}
template<class T> void print(T& a){cout << a<< '\n';}
template<class T,class... Ts> void print(const T&a, const Ts&... b){cout << a;(cout<<...<<(cout<<' ',b));cout<<'\n';}
#endif
#define VI(v,n) vi v(n); input(v)
#define INT(...) int __VA_ARGS__; input(__VA_ARGS__)
#define STR(...) string __VA_ARGS__; input(__VA_ARGS__)
#define CHAR(...) char __VA_ARGS__; input(__VA_ARGS__)
int sign(ll x){return x>0?1:x<0?-1:0;}
ll ceil(ll x,ll y){assert(y!=0);if(sign(x)==sign(y))return (x+y-1)/y;return -((-x/y));}
ll floor(ll x,ll y){assert(y!=0);if(sign(x)==sign(y))return x/y;if(y<0)x*=-1,y*=-1;return x/y-(x%y<0);}
ll abs(ll x,ll y){return abs(x-y);}
ll bit(int n){return 1ll<<n;}
bool ins(string s,string t){return s.find(t)!=string::npos;}
P operator+ (const P &p, const P &q){ return P{p.first+q.first,p.second+q.second};}
P operator- (const P &p, const P &q){ return P{p.first-q.first,p.second-q.second};}
int yesno(bool ok,string y="Yes",string n="No"){ cout<<(ok?y:n)<<endl;return 0;}
int YESNO(bool ok,string y="YES",string n="NO"){ cout<<(ok?y:n)<<endl;return 0;}
int di[]={-1,0,1,0,-1,-1,1,1};
int dj[]={0,1,0,-1,-1,1,-1,1};
const ll INF = 8e18;
using mint = modint1000000007;
//using mint = modint998244353;
//mint stom(const string &s,int b=10){mint res = 0;for(auto c:s)res *= b,res += c-'0';return res;}

signed main() {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    STR(a,b);
    auto solve = [&](string S,int in){
        int n = S.size();
        vector dp(n+1,vector(2,vector(2,vector(3,vector<mint>(8)))));
        dp[0][0][0][0][0] = 1;
        rep(i,n){
            int d = S[i]-'0';
            rep(j,2)rep(k,2)rep(l,3)rep(m,8){
                for(int x=0;x<=(j?9:d);x++){
                    dp[i+1][j|x<d][k|x==3][(l*10+x)%3][(m*10+x)%8]+=dp[i][j][k][l][m];
                }
            }
        }
        mint res = 0;
        rep(k,2)rep(l,3)rep(m,8){
            if(k==0&&l!=0)continue;
            if(m==0)continue;
            res += dp[n][1][k][l][m];
            if(in)res+=dp[n][0][k][l][m];
        }
        return res;
    };
    
    print(solve(b,1)-solve(a,0));
    return 0;
}