#include <bits/stdc++.h>

#define el '\n'

typedef long long ll;
typedef long double ld;

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

using namespace std;

const int N = 5e6 + 5, M = 998244353;

int m, id, cur;
int fact[N], inv[N], dp[23][2], vis[23][2];

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

    return (a + b) % M;
}

int mul(int a, int b) {
    return 1LL * a * b % M;
}

int modPow(int b, int p) {
    if (p == 0)
        return 1;

    int x = modPow(b, p / 2);

    return p % 2 == 0 ? mul(x, x) : mul(b, mul(x, x));
}

int modInvFer(int n) {
    return modPow(n, M - 2);
}

void pre() {
    fact[0] = 1;

    for (int i = 1; i < N; i++)
        fact[i] = mul(fact[i - 1], i);

    inv[N - 1] = modInvFer(fact[N - 1]);

    for (int i = N - 2; i >= 0; i--)
        inv[i] = mul(inv[i + 1], i + 1);
}

int ncr(int n, int r) {
    return mul(fact[n], mul(inv[r], inv[n - r]));
}

int solve(int bit, bool eq) {
    if (bit < 0)
        return 1;

    int &ret = dp[bit][eq];

    if (vis[bit][eq] == id)
        return ret;

    vis[bit][eq] = id;

    ret = 0;

    if (!(cur >> bit & 1))
        ret = solve(bit - 1, eq & !(m >> bit & 1));

    if (!eq || m >> bit & 1)
        ret += solve(bit - 1, eq);

    return ret;
}

void testCase() {
    pre();

    int n, k;
    cin >> n >> k >> m;

    int sol = 0;
    for (int msk = 1; msk <= m; msk++) {
        cur = msk, id++;

        int pc = __builtin_popcount(msk);

        if (pc < k || solve(22, 1) < n)
            continue;

        if ((pc - k) & 1)
            sol = add(sol, -ncr(solve(22, 1), n));
        else
            sol = add(sol, ncr(solve(22, 1), n));
    }

    cout << mul(sol, fact[n]);
}

signed main() {
    Beevo

    int t = 1;
//    cin >> t;

    while (t--)
        testCase();
}