#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

#include <atcoder/all>
#define rep(i, a, b) for (ll i = (ll)(a); i < (ll)(b); i++)
using namespace atcoder;
using namespace std;

typedef long long ll;
using mint = modint1000000007;

mint dp[100010][3][8][2][2];

mint f(string s, bool eq) {
    int n = s.size();
    rep(a, 0, n + 1) rep(b, 0, 3) rep(c, 0, 8) rep(d, 0, 2) rep(e, 0, 2) {
        dp[a][b][c][d][e] = 0;
    }
    dp[0][0][0][0][0] = 1;

    rep(i, 0, n) rep(d, 0, 10) rep(mod3, 0, 3) rep(mod8, 0, 8) rep(in3, 0, 2)
        rep(les, 0, 2) {
        if (les == 0 && d > (s[i] - '0')) continue;
        int nmod3 = (mod3 + d) % 3;
        int nin3 = in3 || (d == 3);
        int nles = les || (d < (s[i] - '0'));
        int p10 = n - 1 - i;
        int nmod8;
        if (p10 >= 3) {
            nmod8 = mod8;
        } else {
            nmod8 = (mod8 + pow_mod(10, p10, 8) * d) % 8;
        }
        dp[i + 1][nmod3][nmod8][nin3][nles] += dp[i][mod3][mod8][in3][les];
    }
    mint ret = 0;
    rep(mod3, 0, 3) rep(mod8, 0, 8) rep(in3, 0, 2) {
        bool aho = mod3 == 0 || in3 == 1;
        bool blue = mod8 == 0;
        if (aho && !blue) {
            ret += dp[n][mod3][mod8][in3][1];
            if (eq) ret += dp[n][mod3][mod8][in3][0];
        }
    }
    return ret;
}

void solve() {
    string s, t;
    cin >> s >> t;
    mint ans = f(t, 1) - f(s, 0);
    cout << ans.val() << endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
}