#include <bits/stdc++.h>

#define el '\n'
#define ll long long
#define ld long double

#define Beevo ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

using namespace std;

const int N = 1e5 + 5, M = 998244353;

string l, r;
int dp[N][11][2][2][2][2];

int add(int a, int b) {
    a = (a % M + M) % M;
    b = (b % M + M) % M;

    return (a + b) % M;
}

int valid(int idx, int d, bool eqL, bool eqR, bool skipL, bool skipR) {
    int valid = 3;

    if (eqL && d < l[idx] - '0' || skipL)
        valid ^= 1;

    if (eqR && d > r[idx] - '0' || skipR)
        valid ^= 2;

    return valid;
}

int solve(int idx, int lst, bool eqL, bool eqR, bool skipL, bool skipR) {
    if (idx >= r.size())
        return (idx == l.size() && !skipL || idx > l.size()) && !skipR;

    int &ret = dp[idx][lst][eqL][eqR][skipL][skipR];

    if (~ret)
        return ret;

    ret = (idx == l.size() && !skipL || idx > l.size());

    for (int d = (idx ? 0 : 1); d < 10; d++) {
        if (d == lst)
            continue;

        int cur = valid(idx, d, eqL, eqR, skipL, skipR);

        ret = add(ret, solve(idx + 1, d, eqL & (l[idx] - '0' == d), eqR & (r[idx] - '0' == d), skipL | !(cur & 1), skipR | !(cur & 2)));
    }

    return ret;
}

void testCase() {
    cin >> l >> r;

    memset(dp, -1, sizeof dp);

    cout << solve(0, 10, 1, 1, 0, 0);
}

signed main() {
    Beevo

    int T = 1;
//    cin >> T;

    while (T--)
        testCase();

    return 0;
}