// #undef _GLIBCXX_DEBUG
// #undef LOCAL

#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
using namespace std;

#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif

#ifdef LOCAL
#include "../algo/debug.hpp"
#else
#define debug(...) void(0)
#endif

constexpr int inf = (1<<30) - 1;
constexpr long long linf = (1LL<<60) - 1;
using ll = long long;
using ld = long double;
#define len(x) int((x).size())
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define LB(c, x) distance(begin(c), lower_bound(all(c), x))
#define UB(c, x) distance(begin(c), upper_bound(all(c), x))
template <typename T, typename S> inline auto min(const T& a, const S& b) { return(a < b ? a : b); }
template <typename T, typename S> inline auto max(const T& a, const S& b) { return(a > b ? a : b); }
template <typename T, typename S> inline bool chmax(T& a, const S& b) { return(a < b ? a = b, 1 : 0); }
template <typename T, typename S> inline bool chmin(T& a, const S& b) { return(a > b ? a = b, 1 : 0); }
// ------------------------------------------------------------------



void solve() {
    string A, B; cin >> A >> B;
    using mint = modint1000000007;
    auto f = [&](string s) -> mint {
        int n = len(s);
        vector dp(n + 1, vector<vector<vector<vector<mint>>>>(2, vector<vector<vector<mint>>>(2, vector<vector<mint>>(3, vector<mint>(8, 0)))));
        dp[0][0][0][0][0] = 1;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < 2; j++) {
                for(int k = 0; k < 2; k++) {
                    for(int l = 0; l < 3; l++) {
                        for(int m = 0; m < 8; m++) {
                            int lim = j ? 9: s[i] - '0';
                            for(int d = 0; d <= lim; d++) {
                                dp[i + 1][j || d < lim][k || d == 3][(l + d)%3][(2*m+d)%8] += dp[i][j][k][l][m];
                            }
                        }
                    }
                }
            }
        }
        mint ans = 0;
        for(int i = 0; i < 2; i++) {
            for(int j = 0; j < 2; j++) {
                for(int k = 0; k < 3; k++) {
                    for(int l = 1; l < 8; l++) {
                        if(j == 1 || k == 0) ans += dp[n][i][j][k][l];
                    }
                }
            }
        }
        return ans;
    };
    A.back() -= 1;
    mint ans = f(B) - f(A);
    cout << ans.val() << endl;
}



int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    // fixed(cout).precision(12);
    int T = 1;
    // cin >> T;
    while(T--) {
        solve();
    }
}