容易发现一个数是萌数当且仅当存在两个相同的数且下标差小于等于 $2$，于是数位 $dp$。

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

const int N = 1010, mod = 1e9 + 7;
int a[N], f[N][20][20];

void init() {
    for (int j = 0; j <= 10; j ++ )
        for (int k = 0; k <= 10; k ++ )
            if (j != k || (j == 10 && k == 10))
                f[0][j][k] = 1;
    for (int i = 1; i <= 1000; i ++ )
        for (int j = 0; j <= 10; j ++ )
            for (int k = 0; k <= 10; k ++ )
                for (int p = 0; p <= 9; p ++ )
                    if ((j != k || (j == 10 && k == 10)) && j != p && p != k)
                        f[i][j][k] = (f[i][j][k] + f[i - 1][k][p]) % mod;
}

int solve(string s) {
    if (s == "0") return 0;
    int n = 0;
    for (int i = s.size() - 1; i >= 0; i -- ) a[++ n] = s[i] - '0';
    while (a[n] == 0 && n > 1) n -- ;
    int lla = 10, la = 10, ans = 0;
    for (int i = n; i >= 1; i -- ) {
        int x = a[i];
        for (int j = (i == n) ? 1 : 0; j < x; j ++ ) {
            if (lla == j && la != j) continue;
            ans = (ans + f[i - 1][la][j]) % mod;// cout << i - 1 << ' ' << la << ' ' << j << ' ' << f[i - 1][la][j] << endl;
        }
        if (i == n) {
            for (int j = 0; j < n - 1; j ++ )
                for (int k = 1; k <= 9; k ++ )
                    ans = (ans + f[j][10][k]) % mod;// cout << j << ' ' << 10 << ' ' << k << ' ' << f[j][10][k]<<endl;
        }
        if (x == la || x == lla) break;
        lla = la, la = x;
        if (i == 1) ans = (ans + 1) % mod;
    }
    return ans;
}

int MOD(string s) {
    int n = 0;
    for (int i = s.size() - 1; i >= 0; i -- ) a[++ n] = s[i] - '0';
    while (a[n] == 0 && n > 1) n -- ;
    int rest = 0;
    for (int i = n; i >= 1; i -- )
        rest = (rest * 10ll + a[i]) % mod;
    return rest;
}

int main() {
    init();
    string l, r;
    cin >> l >> r;
    int pos = l.size() - 1;
    while (l[pos] == '0') l[pos] = '9', pos -- ;
    l[pos] -- ;
    printf("%d\n",((MOD(r) + mod - MOD(l)) % mod + mod - (solve(r) + mod - solve(l)) % mod) % mod);
    return 0;
}