E869120 defined a sequence $a$ like this:


* $a_1=a_2=1$, $a_{k+2}=a_{k+1}+a_k \ (k \ge 1)$


He also defined sequences $d_1, d_2, d_3, \dots , d_n$, as the following recurrence relation :


* $d_{1, j} = a_j$
* $d_{i, j} = \sum_{k = 1}^j d_{i - 1, k} \ (i \ge 2)$


You are given integers $n$ and $m$. Please calculate the value of $d_{n, m}$.
Since the answer can be large number, print the answer modulo $998,244,353$.
Can you solve this problem???

Input

The input is given from standard input in the following format.



> $n \quad m$

Output

* Print $d_{n, m}$ modulo $998,244,353$.



Constraints

* $1 \le n \le 200,000$
* $1 \le m \le 10^{18}$



Subtasks

Subtask 1 [ $100$ points ]


* The testcase in this subtask satisfies $1 \le n, m \le 3,000$.

Subtask 2 [ $170$ points ]


* The testcase in this subtask satisfies $1 \le m \le 200,000$.

Subtask 3 [ $230$ points ]


* The testcase in this subtask satisfies $1 \le n \le 3$.

Subtask 4 [ $420$ points ]


* The testcase in this subtask satisfies $1 \le n \le 1000$.

Subtask 5 [ $480$ points ]


* There are no additional constraints.

Output

* Print $d_{n, m}$ modulo $998,244,353$.



Constraints

* $1 \le n \le 200,000$
* $1 \le m \le 10^{18}$



Subtasks

Subtask 1 [ $100$ points ]


* The testcase in this subtask satisfies $1 \le n, m \le 3,000$.

Subtask 2 [ $170$ points ]


* The testcase in this subtask satisfies $1 \le m \le 200,000$.

Subtask 3 [ $230$ points ]


* The testcase in this subtask satisfies $1 \le n \le 3$.

Subtask 4 [ $420$ points ]


* The testcase in this subtask satisfies $1 \le n \le 1000$.

Subtask 5 [ $480$ points ]


* There are no additional constraints.

Input

The input is given from standard input in the following format.



> $n \quad m$

Examples

Input

4 7


Output

176


Input

12 20


Output

174174144


Input

16 30


Output

102292850
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef double db; 
typedef string str; 

typedef pair<int,int> pi;
typedef pair<ll,ll> pl; 
typedef pair<db,db> pd; 

typedef vector<int> vi; 
typedef vector<ll> vl; 
typedef vector<db> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define mp make_pair
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 998244353;
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; 
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

template<class T> bool ckmin(T& a, const T& b) { 
	return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0; } 
int pct(int x) { return __builtin_popcount(x); } 
int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0 
int fstTrue(function<bool(int)> f, int lo, int hi) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		int mid = (lo+hi)/2; 
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}

// INPUT
template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);

template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }

template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
template<class A> str ts(complex<A> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) { 
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { // containers with begin(), end()
	bool fst = 1; str res = "{";
	for (const auto& x: v) {
		if (!fst) res += ", ";
		fst = 0; res += ts(x);
	}
	res += "}"; return res;
}
template<class A, class B> str ts(pair<A,B> p) {
	return "("+ts(p.f)+", "+ts(p.s)+")"; }

// OUTPUT
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { 
	pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) { 
	pr(h); if (sizeof...(t)) pr(" "); ps(t...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
	cerr << ts(h); if (sizeof...(t)) cerr << ", ";
	DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

// FILE I/O
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { ios_base::sync_with_stdio(0); cin.tie(0); }
void setIO(string s = "") {
	unsyncIO();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

/**
 * Description: modular arithmetic operations 
 * Source: 
	* KACTL
	* https://codeforces.com/blog/entry/63903
	* https://codeforces.com/contest/1261/submission/65632855 (tourist)
	* https://codeforces.com/contest/1264/submission/66344993 (ksun)
 * Verification: 
	* https://open.kattis.com/problems/modulararithmetic
 */

struct mi {
	typedef decay<decltype(MOD)>::type T; 
 	/// don't silently convert to T
	T v; explicit operator T() const { return v; }
	mi() { v = 0; }
	mi(ll _v) { 
		v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
		if (v < 0) v += MOD;
	}
	friend bool operator==(const mi& a, const mi& b) { 
		return a.v == b.v; }
	friend bool operator!=(const mi& a, const mi& b) { 
		return !(a == b); }
	friend bool operator<(const mi& a, const mi& b) { 
		return a.v < b.v; }
	friend void re(mi& a) { ll x; re(x); a = mi(x); }
	friend str ts(mi a) { return ts(a.v); }
   
	mi& operator+=(const mi& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	mi& operator-=(const mi& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	mi& operator*=(const mi& m) { 
		v = (ll)v*m.v%MOD; return *this; }
	mi& operator/=(const mi& m) { return (*this) *= inv(m); }
	friend mi pow(mi a, ll p) {
		mi ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans;
	}
	friend mi inv(const mi& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mi operator-() const { return mi(-v); }
	mi& operator++() { return *this += 1; }
	mi& operator--() { return *this -= 1; }
	friend mi operator+(mi a, const mi& b) { return a += b; }
	friend mi operator-(mi a, const mi& b) { return a -= b; }
	friend mi operator*(mi a, const mi& b) { return a *= b; }
	friend mi operator/(mi a, const mi& b) { return a /= b; }
};
typedef vector<mi> vmi;
typedef pair<mi,mi> pmi;
typedef vector<pmi> vpmi;

vector<vmi> scmb; // small combinations
void genComb(int SZ) {
	scmb.assign(SZ,vmi(SZ)); scmb[0][0] = 1;
	FOR(i,1,SZ) F0R(j,i+1) 
		scmb[i][j] = scmb[i-1][j]+(j?scmb[i-1][j-1]:0);
}

/**
 * Description: Basic poly ops including division. Can replace \texttt{T} with double, complex.
 * Source: Own. Also see
 	* https://github.com/kth-competitive-programming/kactl/blob/master/content/numerical/PolyInterpolate.h
 	* https://github.com/ecnerwala/icpc-book/blob/master/content/numerical/fft.cpp
 * Verification: see FFT
 */

// #include "../../number-theory (11.1)/Modular Arithmetic/ModInt.h"

typedef mi T; using poly = vector<T>;
void remz(poly& p) { while (sz(p)&&p.bk==T(0)) p.pop_back(); }
poly rev(poly p) { reverse(all(p)); return p; }
poly shift(poly p, int x) { p.insert(begin(p),x,0); return p; }
poly RSZ(poly p, int x) { p.rsz(x); return p; }
T eval(const poly& p, T x) { // evaluate at point x
	T res = 0; R0F(i,sz(p)) res = x*res+p[i]; 
	return res; }
poly dif(const poly& p) { // differentiate
	poly res; FOR(i,1,sz(p)) res.pb(T(i)*p[i]); 
	return res; }
poly integ(const poly& p) { // integrate
	poly res(sz(p)+1); F0R(i,sz(p)) res[i+1] = p[i]/T(i+1);
	return res; }

poly& operator+=(poly& l, const poly& r) {
	l.rsz(max(sz(l),sz(r))); F0R(i,sz(r)) l[i] += r[i]; 
	return l; }
poly& operator-=(poly& l, const poly& r) {
	l.rsz(max(sz(l),sz(r))); F0R(i,sz(r)) l[i] -= r[i]; 
	return l; }
poly& operator*=(poly& l, const T& r) { trav(t,l) t *= r; 
	return l;	 }
poly& operator/=(poly& l, const T& r) { trav(t,l) t /= r; 
	return l; }
poly operator+(poly l, const poly& r) { return l += r; }
poly operator-(poly l, const poly& r) { return l -= r; }
poly operator-(poly l) { trav(t,l) t *= -1; return l; }
poly operator*(poly l, const T& r) { return l *= r; }
poly operator*(const T& r, const poly& l) { return l*r; }
poly operator/(poly l, const T& r) { return l /= r;	 }
poly operator*(const poly& l, const poly& r) {
	if (!min(sz(l),sz(r))) return {};
	poly x(sz(l)+sz(r)-1); 
	F0R(i,sz(l)) F0R(j,sz(r)) x[i+j] += l[i]*r[j];
	return x;
}
poly& operator*=(poly& l, const poly& r) { return l = l*r; }

pair<poly,poly> quoRem(poly a, poly b) { 
	remz(a); remz(b); assert(sz(b));
	auto B = T(1)/b.bk; trav(t,a) t *= B; trav(t,b) t *= B;
	poly q(max(sz(a)-sz(b)+1,0));
	for (int dif; (dif=sz(a)-sz(b)) >= 0; remz(a)) {
		q[dif] = a.bk; F0R(i,sz(b)) a[i+dif] -= q[dif]*b[i]; }
	return {q,a}; // quotient, remainder
}
/**
poly quo(poly a, poly b) { return quoRem(a,b).f; }
poly rem(poly a, poly b) { return quoRem(a,b).s; }

poly a = {1,3,5,8,6,0,0,0,0}, b = {1,5,1};
ps(quoRem(a,b)); a = 2*a, b = 2*b; ps(quoRem(a,b));
*/

/**
 * Description: pre-compute factorial mod inverses,
 	* assumes $MOD$ is prime and $SZ < MOD$.
 * Time: O(SZ)
 * Source: KACTL
 * Verification: https://dmoj.ca/problem/tle17c4p5
 */

vi invs, fac, ifac;
void genFac(int SZ) {
	invs.rsz(SZ), fac.rsz(SZ), ifac.rsz(SZ); 
	invs[1] = fac[0] = ifac[0] = 1; 
	FOR(i,2,SZ) invs[i] = MOD-(ll)MOD/i*invs[MOD%i]%MOD;
	FOR(i,1,SZ) {
		fac[i] = (ll)fac[i-1]*i%MOD;
		ifac[i] = (ll)ifac[i-1]*invs[i]%MOD;
	}
}

ll comb(int a, int b) {
	if (a < b || b < 0) return 0;
	return (ll)fac[a]*ifac[b]%MOD*ifac[a-b]%MOD;
}

vmi getCoef(ll m) {
	if (m == 1) return {0,1};
	if (m&1) {
		vmi v = getCoef(m-1);
		return quoRem(v*poly{0,1},poly{-1,-1,1}).s;
	}
	vmi v = getCoef(m/2);
	return quoRem(v*v,poly{-1,-1,1}).s;
}

mi com = 1;
int extra;

void ad(ll a) {
	if (a%MOD == 0) extra ++;
	else com *= a;
}

void del(ll a) {
	if (a%MOD == 0) extra --;
	else com /= a;
}

ll n,m;

int main() {
	setIO(); re(n,m);
	genFac(MX);
	vmi fib = {0,1};
	FOR(i,2,2*n+5) fib.pb(fib[i-1]+fib[i-2]);
	// A(x)/(1-x)^{n-1}
	// B(x)/(1-x-x^2)
	// first find B(x)
	// 0,1,1,2,3,5,8,13
	poly B;
	if (n == 1) B = {0,1};
	else B = {fib[2*n-2],fib[2*n-3]};
	//B *= -1;
	//if (n%2 == 0) B *= -1;
	vmi coef;
	F0R(i,n) {
		mi tmp = comb(n-1,i); if (i&1) tmp *= -1;
		coef.pb(tmp);
	}
	poly num = poly{0,1}-B*coef;
	poly A = quoRem(num,poly{1,-1,-1}).f;
	dbg(getCoef(2));
	poly huh = getCoef(m);
	poly dem = {-1,-1,1};
	poly v(8); v[7] = 1;
	dbg(quoRem(v,dem).s);
	dbg(m);
	//dbg("OH",huh);
	// 
	//dbg("OH",huh,B);
	mi ans = B[0]*huh[0]+huh[1]*(B[0]+B[1]);
	dbg(A);
	dbg(B);
	//dbg("??",A*poly{1,-1,-1}+B*coef);
	//dbg("AA",ans);
	FOR(i,1,n-1) ad(m+i);
	FOR(i,1,n-1) del(i);
	if (n > 1) {
		F0R(i,min((ll)sz(A),m+1)) {
			ans += (extra?0:com)*A[i];
			del(m-i+n-2);
			ad(m-i);
		}
	}
	// A(x)/(1-x)^{n-1}
	// B(x)/(1-x-x^2)
	// dbg(num,A,B);
	ps(ans);
	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
*/

