容易想到建图，即 $i \to a_i$。
容易发现环上的点数值相同，所以容易想到缩点。
容易发现图变成 DAG，准确地说是森林，所以容易想到拓扑。
由于 $N,M \leq 2025$，所以容易想到 $n^2$ DP，设 $dp_{i,j}$ 表示 $i$ 子树处理完，$i$ 填 $j$ 的方案数，转移是容易的。
于是这题就容易地做完了。

#include <bits/stdc++.h>
using namespace std;
const int N = 2050, M = N << 1, mod = 998244353;
int n, m, to[N];
int dfn[N], low[N], tot = 0;
int stk[N], top = 0;
bool in_stk[N];
int scc[N], cnt = 0;

struct Graph {
    int h[N], e[M], ne[M], idx = 0;
    void init() {
        for (int i = 0; i < N; i++) h[i] = -1;
        idx = 0;
    }
    void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }

    void tarjan(int u) {
        dfn[u] = low[u] = ++tot;
        stk[++top] = u, in_stk[u] = 1;
        for (int i = h[u]; ~i; i = ne[i]) {
            int v = e[i];
            if (!dfn[v]) {
                tarjan(v);
                low[u] = min(low[u], low[v]);
            } else if (in_stk[v]) {
                low[u] = min(low[u], dfn[v]);
            }
        }
        if (low[u] == dfn[u]) {
            ++cnt;
            while (stk[top] != u) {
                scc[stk[top]] = cnt;
                in_stk[stk[top]] = 0;
                top--;
            }
            scc[stk[top]] = cnt;
            in_stk[stk[top]] = 0;
            top--;
        }
    }
} G;

int h[N], e[M], ne[M], idx = 0;
int in[N], out[N];
void add(int a, int b) {
    in[b]++, out[a]++;
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

queue<int> q;
long long dp[N][N], sum[N][N], ans = 0;
void topsort() {
    for (int i = 1; i <= cnt; i++) {
        for (int j = 1; j <= m; j++) dp[i][j] = 1, sum[i][j] = 0;
        if (!in[i]) q.push(i);
    }
    while (q.size()) {
        int u = q.front(); q.pop();
        for (int j = 1; j <= m; j++) sum[u][j] = (sum[u][j - 1] + dp[u][j]) % mod;
        for (int i = h[u]; ~i; i = ne[i]) {
            int v = e[i];
            for (int j = 1; j <= m; j++) dp[v][j] = (dp[v][j] * sum[u][j]) % mod; 
            if (--in[v] == 0) q.push(v);
        }
    }
    ans = 1ll;
    for (int i = 1; i <= cnt; i++) {
        if (out[i]) continue;
        ans = (ans * sum[i][m]) % mod;
    }
}

int main() {
    G.init();
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &to[i]), G.add(i, to[i]);
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) G.tarjan(i);
    for (int i = 1; i <= cnt; i++) h[i] = -1; idx = 0;
    for (int u = 1; u <= n; u++) {
        int v = to[u];
        if (scc[u] == scc[v]) continue;
        add(scc[u], scc[v]);
    }
    topsort();
    printf("%lld\n", ans);
    return 0;
}