#include #include using namespace std; using namespace atcoder; //using mint = modint1000000007; //const int mod = 1000000007; using mint = modint998244353; const int mod = 998244353; //const int INF = 1e9; //const long long LINF = 1e18; #define rep(i, n) for (int i = 0; i < (n); ++i) #define rep2(i,l,r)for(int i=(l);i<(r);++i) #define rrep(i, n) for (int i = (n-1); i >= 0; --i) #define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i) #define all(x) (x).begin(),(x).end() #define allR(x) (x).rbegin(),(x).rend() #define endl "\n" #define P pair template inline bool chmax(A& a, const B& b) { if (a < b) { a = b; return true; } return false; } template inline bool chmin(A& a, const B& b) { if (a > b) { a = b; return true; } return false; } // combination mod prime // https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619 struct combination { vector fact, ifact; combination(int n) :fact(n + 1), ifact(n + 1) { assert(n < mod); fact[0] = 1; for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i; ifact[n] = fact[n].inv(); for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i; } mint operator()(int n, int k) { return com(n, k); } mint com(int n, int k) { //負の二項係数を考慮する場合にコメントアウトを外す //if (n < 0) return com(-n, k) * (k % 2 ? -1 : 1); if (k < 0 || k > n) return 0; return fact[n] * ifact[k] * ifact[n - k]; } mint inv(int n, int k) { //if (n < 0) return inv(-n, k) * (k % 2 ? -1 : 1); if (k < 0 || k > n) return 0; return ifact[n] * fact[k] * fact[n - k]; } mint p(int n, int k) { return fact[n] * ifact[n - k]; } }com(2000006); mapmp; mint dp(int n, long long b, long long c) { if (b < c)return 0; if (b < 0)return 0; if (mp.count({ b,c }))return mp[{b, c}]; if (0 == b)return 0; mint ret = 0; rep(j, n + 1) { if (j > b)break; if (b % 2 != j % 2)continue; if (c % 2 != j % 2)continue; long long nb = (b - j) / 2; long long nc = (c ^ (j % 2)) / 2; ret += dp(n, nb, nc) * com(n, j); } return mp[{b, c}] = ret; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n; cin >> n; long long b, c; cin >> b >> c; mp[{0, 0}] = 1; mint ans = dp(n, b, c); cout << ans.val() << endl; return 0; }