// Sea, Your Memory
#include<bits/stdc++.h>
#define LL long long
#define DB double
#define MOD 998244353
#define ls(x) x << 1
#define rs(x) x << 1 | 1
#define lowbit(x) x & (-x)
#define PII pair<int, int>
#define MP make_pair
#define VI vector<int>
#define VII vector<int>::iterator
#define all(x) x.begin(), x.end()
#define EB emplace_back
#define SI set<int>
#define SII set<int>::iterator
#define QI queue<int>
#define fi first
#define se second
using namespace std;
template<typename T> void chkmn(T &a, const T &b) { (a > b) && (a = b); }
template<typename T> void chkmx(T &a, const T &b) { (a < b) && (a = b); }
int inc(const int &a, const int &b) { return a + b >= MOD ? a + b - MOD : a + b; }
int dec(const int &a, const int &b) { return a - b < 0 ? a - b + MOD : a - b; }
int mul(const int &a, const int &b) { return 1LL * a * b % MOD; }
int sqr(const int &a) { return 1LL * a * a % MOD; }
void Inc(int &a, const int &b) { ((a += b) >= MOD) && (a -= MOD); }
void Dec(int &a, const int &b) { ((a -= b) < 0) && (a += MOD); }
void Mul(int &a, const int &b) { a = 1LL * a * b % MOD; }
void Sqr(int &a) { a = 1LL * a * a % MOD; }
int qwqmi(int x, int k = MOD - 2)
{
    int res = 1;
    while(k)
    {
        if(k & 1) Mul(res, x);
        Sqr(x), k >>= 1;
    }
    return res;
}
template<typename T> void read(T &x)
{
	x = 0;
	int f = 1;
	char ch = getchar();
	while(!isdigit(ch))
	{
		if(ch == '-')
			f = -1;
		ch = getchar();
	}
	while(isdigit(ch))
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	x = x * f;
}
const int N = 3e5 + 5;
const int B = 800;
const int B2 = 2 * B + 5;
const int _B = 2 * N / B;
const int M = _B * _B;
int n, C;
int f[B2][B2];
int Solve1()
{
	for(int i = 1; i < B; ++i)
		f[i][i] = 1;
	for(int s = 1; s <= n; ++s)
	{
		for(int x = 1; x <= 2 * B && s - x > 0; ++x)
		{
			int t = s - x, lst = t % B2, now = s % B2;
			f[now][x] = inc(mul(C, f[lst][x + 1]), f[lst][x - 1]);
		}
	}
	int res = 0;
	for(int x = 1; x <= 2 * B; ++x)
		Inc(res, f[n % B2][x]);
//	cerr << "res1 : " << res << '\n';
	return res;
}
int g[2][M * 2 + 5];
int Solve2()
{
	int now = 1, lst = 0;
	g[1][M] = 1;
	int res = (n >= B);
	for(int i = 2; i <= _B; ++i)
	{
		swap(now, lst);
		memset(g[now], 0, sizeof(g[now]));
		for(int j = -(i - 1) * (i - 2) / 2; j <= (i - 1) * (i - 2) / 2; ++j)
		{
			if(!g[lst][j + M]) continue;
			Inc(g[now][j + (i - 1) + M], g[lst][j + M]);
			Inc(g[now][j - (i - 1) + M], mul(C, g[lst][j + M]));
		}
		for(int j = -i * (i - 1) / 2; j <= i * (i - 1) / 2; ++j)
			if((n - j) % i == 0 && (n - j) / i >= B)
				Inc(res, g[now][j + M]);
	}
//	cerr << "res2 : " << res << '\n';
	return res;
}
int main()
{
	read(n), read(C);
	int ans = inc(Solve1(), Solve2());
	printf("%d\n", ans);
    return 0;
}