Problem: We have an N \times M grid. The square at the i-th row and j-th column will be denoted as (i,j). Particularly, the top-left square will be denoted as (1,1), and the bottom-right square will be denoted as (N,M). Takahashi painted some of the squares (possibly zero) black, and painted the other squares white.

We will define an integer sequence A of length N, and two integer sequences B and C of length M each, as follows:

* A_i(1\leq i\leq N) is the minimum j such that (i,j) is painted black, or M+1 if it does not exist.
* B_i(1\leq i\leq M) is the minimum k such that (k,i) is painted black, or N+1 if it does not exist.
* C_i(1\leq i\leq M) is the maximum k such that (k,i) is painted black, or 0 if it does not exist.



How many triples (A,B,C) can occur? Find the count modulo 998244353.

Constraints

* 1 \leq N \leq 8000
* 1 \leq M \leq 200
* N and M are integers.



Partial Score

* 1500 points will be awarded for passing the test set satisfying N\leq 300.

Constraints

* 1 \leq N \leq 8000
* 1 \leq M \leq 200
* N and M are integers.

Input

Input is given from Standard Input in the following format:


N M

Output

Print the number of triples (A,B,C), modulo 998244353.

Examples

Input

2 3


Output

64


Input

4 3


Output

2588


Input

17 13


Output

229876268


Input

5000 100


Output

57613837

Solution (in CPP):

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

/**
 * Author: Andrew He
 * Description: FFT/NTT, polynomial mod/log/exp
 * Source: http://neerc.ifmo.ru/trains/toulouse/2017/fft2.pdf
 * Papers about accuracy: http://www.daemonology.net/papers/fft.pdf, http://www.cs.berkeley.edu/~fateman/papers/fftvsothers.pdf
 * For integers rounding works if $(|a| + |b|)\max(a, b) < \mathtt{\sim} 10^9$, or in theory maybe $10^6$.
 */

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define per(i, a, b) for(int i = (b)-1; i >= (a); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

namespace fft {
#if FFT
// FFT
using dbl = double;
struct num { /// start-hash
	dbl x, y;
	num(dbl x_ = 0, dbl y_ = 0) : x(x_), y(y_) { }
};
inline num operator+(num a, num b) { return num(a.x + b.x, a.y + b.y); }
inline num operator-(num a, num b) { return num(a.x - b.x, a.y - b.y); }
inline num operator*(num a, num b) { return num(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); }
inline num conj(num a) { return num(a.x, -a.y); }
inline num inv(num a) { dbl n = (a.x*a.x+a.y*a.y); return num(a.x/n,-a.y/n); }
/// end-hash
#else
// NTT
const int mod = 998244353, g = 3;
// For p < 2^30 there is also (5 << 25, 3), (7 << 26, 3),
// (479 << 21, 3) and (483 << 21, 5). Last two are > 10^9.
struct num { /// start-hash
	int v;
	num(ll v_ = 0) : v(int(v_ % mod)) { if (v<0) v+=mod; }
	explicit operator int() const { return v; }
};
inline num operator+(num a,num b){return num(a.v+b.v);}
inline num operator-(num a,num b){return num(a.v+mod-b.v);}
inline num operator*(num a,num b){return num(1ll*a.v*b.v);}
inline num pow(num a, int b) {
	num r = 1;
	do{if(b&1)r=r*a;a=a*a;}while(b>>=1);
	return r;
}
inline num inv(num a) { return pow(a, mod-2); }
/// end-hash
#endif

using vn = vector<num>;
vi rev({0, 1});
vn rt(2, num(1)), fa, fb;

inline void init(int n) { /// start-hash
	if (n <= sz(rt)) return;
	rev.resize(n);
	rep(i,0,n) rev[i] = (rev[i>>1] | ((i&1)*n)) >> 1;
	rt.reserve(n);
	for (int k = sz(rt); k < n; k *= 2) {
		rt.resize(2*k);
#if FFT
		double a=M_PI/k; num z(cos(a),sin(a)); // FFT
#else
		num z = pow(num(g), (mod-1)/(2*k)); // NTT
#endif
		rep(i,k/2,k) rt[2*i] = rt[i], rt[2*i+1] = rt[i]*z;
	}
} /// end-hash

inline void fft(vector<num> &a, int n) { /// start-hash
	init(n);
	int s = __builtin_ctz(sz(rev)/n);
	rep(i,0,n) if (i < rev[i]>>s) swap(a[i], a[rev[i]>>s]);
	for (int k = 1; k < n; k *= 2)
		for (int i = 0; i < n; i += 2 * k) rep(j,0,k) {
			num t = rt[j+k] * a[i+j+k];
			a[i+j+k] = a[i+j] - t;
			a[i+j] = a[i+j] + t;
		}
} /// end-hash

// Complex/NTT
vn multiply(vn a, vn b) { /// start-hash
	int s = sz(a) + sz(b) - 1;
	if (s <= 0) return {};
	int L = s > 1 ? 32 - __builtin_clz(s-1) : 0, n = 1 << L;
	a.resize(n), b.resize(n);
	fft(a, n);
	fft(b, n);
	num d = inv(num(n));
	rep(i,0,n) a[i] = a[i] * b[i] * d;
	reverse(a.begin()+1, a.end());
	fft(a, n);
	a.resize(s);
	return a;
} /// end-hash

// Complex/NTT power-series inverse
// Doubles b as b[:n] = (2 - a[:n] * b[:n/2]) * b[:n/2]
vn inverse(const vn& a) { /// start-hash
	if (a.empty()) return {};
	vn b({inv(a[0])});
	b.reserve(2*a.size());
	while (sz(b) < sz(a)) {
		int n = 2*sz(b);
		b.resize(2*n, 0);
		if (sz(fa) < 2*n) fa.resize(2*n);
		fill(fa.begin(), fa.begin()+2*n, 0);
		copy(a.begin(), a.begin()+min(n,sz(a)), fa.begin());
		fft(b, 2*n);
		fft(fa, 2*n);
		num d = inv(num(2*n));
		rep(i, 0, 2*n) b[i] = b[i] * (2 - fa[i] * b[i]) * d;
		reverse(b.begin()+1, b.end());
		fft(b, 2*n);
		b.resize(n);
	}
	b.resize(a.size());
	return b;
} /// end-hash

#if FFT
// Double multiply (num = complex)
using vd = vector<double>;
vd multiply(const vd& a, const vd& b) { /// start-hash
	int s = sz(a) + sz(b) - 1;
	if (s <= 0) return {};
	int L = s > 1 ? 32 - __builtin_clz(s-1) : 0, n = 1 << L;
	if (sz(fa) < n) fa.resize(n);
	if (sz(fb) < n) fb.resize(n);

	fill(fa.begin(), fa.begin() + n, 0);
	rep(i,0,sz(a)) fa[i].x = a[i];
	rep(i,0,sz(b)) fa[i].y = b[i];
	fft(fa, n);
	trav(x, fa) x = x * x;
	rep(i,0,n) fb[i] = fa[(n-i)&(n-1)] - conj(fa[i]);
	fft(fb, n);
	vd r(s);
	rep(i,0,s) r[i] = fb[i].y / (4*n);
	return r;
} /// end-hash

// Integer multiply mod m (num = complex) /// start-hash
vi multiply_mod(const vi& a, const vi& b, int m) {
	int s = sz(a) + sz(b) - 1;
	if (s <= 0) return {};
	int L = s > 1 ? 32 - __builtin_clz(s-1) : 0, n = 1 << L;
	if (sz(fa) < n) fa.resize(n);
	if (sz(fb) < n) fb.resize(n);

	rep(i,0,sz(a)) fa[i] = num(a[i] & ((1<<15)-1), a[i] >> 15);
	fill(fa.begin()+sz(a), fa.begin() + n, 0);
	rep(i,0,sz(b)) fb[i] = num(b[i] & ((1<<15)-1), b[i] >> 15);
	fill(fb.begin()+sz(b), fb.begin() + n, 0);

	fft(fa, n);
	fft(fb, n);
	double r0 = 0.5 / n; // 1/2n
	rep(i,0,n/2+1) {
		int j = (n-i)&(n-1);
		num g0 = (fb[i] + conj(fb[j])) * r0;
		num g1 = (fb[i] - conj(fb[j])) * r0;
		swap(g1.x, g1.y); g1.y *= -1;
		if (j != i) {
			swap(fa[j], fa[i]);
			fb[j] = fa[j] * g1;
			fa[j] = fa[j] * g0;
		}
		fb[i] = fa[i] * conj(g1);
		fa[i] = fa[i] * conj(g0);
	}
	fft(fa, n);
	fft(fb, n);
	vi r(s);
	rep(i,0,s) r[i] = int((ll(fa[i].x+0.5)
				+ (ll(fa[i].y+0.5) % m << 15)
				+ (ll(fb[i].x+0.5) % m << 15)
				+ (ll(fb[i].y+0.5) % m << 30)) % m);
	return r;
} /// end-hash
#endif

} // namespace fft




using ll = long long;
ll modinv(ll a, ll m) {
	assert(m > 0);
	if (m == 1) return 0;
	a %= m;
	if (a < 0) a += m;
	assert(a != 0);
	if (a == 1) return 1;
	return m - modinv(m, a) * m / a;
}

template <int MOD_> struct modnum {
private:
	int v;
public:
	static const int MOD = MOD_;

	modnum() : v(0) {}
	modnum(ll v_) : v(int(v_ % MOD)) { if (v < 0) v += MOD; }
	explicit operator int () const { return v; }
	friend bool operator == (const modnum& a, const modnum& b) { return a.v == b.v; }
	friend bool operator != (const modnum& a, const modnum& b) { return a.v != b.v; }

	modnum operator ~ () const {
		modnum res;
		res.v = modinv(v, MOD);
		return res;
	}

	modnum& operator += (const modnum& o) {
		v += o.v;
		if (v >= MOD) v -= MOD;
		return *this;
	}
	modnum& operator -= (const modnum& o) {
		v -= o.v;
		if (v < 0) v += MOD;
		return *this;
	}
	modnum& operator *= (const modnum& o) {
		v = int(ll(v) * ll(o.v) % MOD);
		return *this;
	}
	modnum& operator /= (const modnum& o) {
		return *this *= (~o);
	}

	friend modnum operator + (const modnum& a, const modnum& b) { return modnum(a) += b; }
	friend modnum operator - (const modnum& a, const modnum& b) { return modnum(a) -= b; }
	friend modnum operator * (const modnum& a, const modnum& b) { return modnum(a) *= b; }
	friend modnum operator / (const modnum& a, const modnum& b) { return modnum(a) /= b; }
};

using num = modnum<998244353>;

vector<num> mult(const vector<num> &_a, const vector<num> &_b){
	vector<fft::num> a, b;
	for(num x : _a) a.push_back(int(x));
	for(num x : _b) b.push_back(int(x));
	vector<fft::num> c = fft::multiply(a, b);
	vector<num> res;
	for(fft::num x : c) res.push_back(int(x));
	return res;
}

vector<num> fact;
vector<num> ifact;

int main(){
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n, m;
	cin >> n >> m;

	fact = {1};
	for(int i = 1; i < 100000; i++) fact.push_back(i * fact[i-1]);
	for(num x : fact) ifact.push_back(1 / x);

	vector<num> dp(n+1, 0);
	for(int p = 0; p <= n; p++){
		dp[p] += fact[n] * ifact[p] * ifact[n-p];
	}
	for(int _ = 0; _ < m; _++){
		vector<num> r1(n+1, 0);
		vector<num> r2(n+1, 0);
		for(int p = 0; p <= n; p++){
			r1[p] = dp[p] * fact[p];
		}
		for(int a = 2; a <= n; a++){
			r2[n-a] += ifact[a];
		}
		vector<num> r = mult(r1, r2);
		r.erase(r.begin(), r.begin() + n);
		for(int i = 0; i <= n; i++){

			r[i] *= ifact[i];
		}
		for(int i = n; i >= 0; i--){
			if(i > 0) r[i] += 2 * r[i-1];
			if(i > 1) r[i] += r[i-2];
		}

		for(int p = 0; p <= n; p++){
			// empty
			r[p] += dp[p];
			// one, not highest
			r[p] += p * dp[p];
			if(p) r[p-1] += p * dp[p];
		}
		dp = r;
	}
	num ans = dp[0];
	cout << (int)ans << '\n';
}