考虑我们如何判定一个排列是否能成为最终答案。连边 $i \to p_i$，设环数为 $k$，那么最少交换次数为 $n - k$。那么充要条件是，每个环所有点的 $c_i$ 相同，并且 $n - k \le K$ 且 $2 \mid (K - (n - k))$。$K$ 和 $n - k$ 奇偶性相同是因为，如果交换 $p_i, p_j$，其中 $i \ne j$，那么环数恰好改变 $1$。这个的证明是平凡的，考虑它们原本在或者不在同一个环内即可。

也就是说现在要求出，每个环所有点的 $c_i$ 相同，共形成了 $k$ 个环的 $p_i$ 的方案数。考虑一个 $O(n^2)$ 的 dp，$f_{i, j}$ 表示，考虑了 $1 \sim i$，共形成了 $j$ 个环的方案数。转移就是讨论 $i$ 这个点是自环还是加到之前的环里面。设 $a_i = \sum\limits_{j = 1}^{i - 1} [c_j = c_i]$，于是有：

$$f_{i, j} = f_{i - 1, j - 1} + f_{i - 1, j} \times a_i$$

考虑优化。发现经过一次转移，$f_{i, j}$ 可能转移到 $f_{i + 1, j}$ 和 $f_{i + 1, j + 1}$。考虑构造多项式 $g(x) = \prod\limits_{i = 1}^n (a_i + x)$，就是要求 $g(x)$ 的第 $1 \sim n$ 次项。直接乘起来复杂度不对，考虑一个分治，分治区间 $[l, r]$ 表示计算 $\prod\limits_{i = l}^r (a_i + x)$。设 $mid = \left\lfloor\frac{l + r}{2}\right\rfloor$，把 $[l, mid]$ 和 $[mid + 1, r]$ 的结果乘起来即可。时间复杂度 $O(n \log^2 n)$。

// Problem: Ex - Rearranging Problem
// Contest: AtCoder - AtCoder Beginner Contest 247
// URL: https://atcoder.jp/contests/abc247/tasks/abc247_h
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 1000100;
const ll mod = 998244353, G = 3;

inline ll qpow(ll b, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1) {
			res = res * b % mod;
		}
		b = b * b % mod;
		p >>= 1;
	}
	return res;
}

ll n, m, a[maxn], b[maxn], r[maxn];

typedef vector<ll> poly;

inline poly NTT(poly a, int op) {
	int n = (int)a.size();
	for (int i = 0; i < n; ++i) {
		if (i < r[i]) {
			swap(a[i], a[r[i]]);
		}
	}
	for (int k = 1; k < n; k <<= 1) {
		ll wn = qpow(op == 1 ? G : qpow(G, mod - 2), (mod - 1) / (k << 1));
		for (int i = 0; i < n; i += (k << 1)) {
			ll w = 1;
			for (int j = 0; j < k; ++j, w = w * wn % mod) {
				ll x = a[i + j], y = w * a[i + j + k] % mod;
				a[i + j] = (x + y) % mod;
				a[i + j + k] = (x - y + mod) % mod;
			}
		}
	}
	if (op == -1) {
		ll inv = qpow(n, mod - 2);
		for (int i = 0; i < n; ++i) {
			a[i] = a[i] * inv % mod;
		}
	}
	return a;
}

inline poly operator * (poly a, poly b) {
	a = NTT(a, 1);
	b = NTT(b, 1);
	int n = (int)a.size();
	for (int i = 0; i < n; ++i) {
		a[i] = a[i] * b[i] % mod;
	}
	a = NTT(a, -1);
	return a;
}

poly dfs(int l, int r) {
	if (l == r) {
		return (poly){a[l], 1};
	}
	int mid = (l + r) >> 1;
	poly L = dfs(l, mid), R = dfs(mid + 1, r);
	int k = 0;
	while ((1 << k) <= r - l + 1) {
		++k;
	}
	for (int i = 1; i < (1 << k); ++i) {
		::r[i] = (::r[i >> 1] >> 1) | ((i & 1) << (k - 1));
	}
	poly A(1 << k), B(1 << k);
	for (int i = 0; i <= mid - l + 1; ++i) {
		A[i] = L[i];
	}
	for (int i = 0; i <= r - mid; ++i) {
		B[i] = R[i];
	}
	poly res = A * B;
	res.resize(r - l + 2);
	return res;
}

void solve() {
	scanf("%lld%lld", &n, &m);
	for (int i = 1, x; i <= n; ++i) {
		scanf("%d", &x);
		a[i] = (b[x]++);
	}
	poly res = dfs(1, n);
	ll ans = 0;
	for (int i = 1; i <= n; ++i) {
		if (n - i <= m && (m - (n - i)) % 2 == 0) {
			ans = (ans + res[i]) % mod;
		}
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}