#include <bits/stdc++.h>
#ifdef LOCAL
#include "debug.h"
#else
#define deb(...)
#endif
using namespace std;
#define ll long long
//#define int long long
#define ull unsigned long long
#define pii pair<int, int>
#define db double
#define baoliu(x, y) cout << fixed << setprecision(y) << x
#define endl "\n"
#define alls(x) (x).begin(), (x).end()
#define fs first
#define sec second
#define bug(x) cerr << #x << " = " << x << endl
const int N = 2e5 + 10;
const int M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1.0);
int n, m;
int a[N];
struct PAM {
    static constexpr int asz = 12;
    struct Node {
        int len;
        int fail;
        int dep;      // 以这个节点结尾的回文子串的数量(回文fail树的深度)
        int cnt = 0;  // 同样的回文结构出现次数
        array<int, asz> next;
        Node() : len{}, fail{}, dep{}, next{} {}
    };
    vector<Node> t;
    vector<int> idpos;
    int last;
    string s;
    PAM() {
        init();
    }
    void init() {
        t.assign(2, Node());
        t[0].len = -1;  // 0：奇根
        last = 1;       // 1：偶根
        s.clear();
    }
    int newNode() {
        t.emplace_back();  // Node()
        return t.size() - 1;
    }

    bool add(char c, char offset = '0') {
        int pos = s.size();
        s += c;
        int ch = c - offset;
        int cur = last, curlen = 0;

        while (true) {
            curlen = t[cur].len;
            if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])
                break;
            cur = t[cur].fail;
        }  // 找到在哪个节点后面建新点
        if (t[cur].next[ch]) {
            last = t[cur].next[ch];
            idpos.push_back(last);
            t[last].cnt += 1;
            return false;
        }

        int num = newNode();
        last = num;
        idpos.push_back(last);
        t[num].len = t[cur].len + 2;
        t[cur].next[ch] = num;

        if (t[num].len == 1) {  // 如果为单字符，指向偶根
            t[num].fail = 1;
            t[num].dep = 1;
            t[num].cnt = 1;
            return true;
        }

        while (true) {  // 为新节点找fail，从父亲的fail开始找
            cur = t[cur].fail;
            curlen = t[cur].len;
            if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) {
                t[num].fail = t[cur].next[ch];
                break;
            }
        }
        t[num].cnt = 1;
        t[num].dep = 1 + t[t[num].fail].dep;

        return true;
    }
    int tot = 0;
    void work(string tt) {
        for (auto x : tt) add(x);
        tot = t.size() - 1;
        for (int i = tot; i >= 0; i--) {
            int fa = t[i].fail;
            t[fa].cnt += t[i].cnt;
        }
    }
};

PAM pam1;
PAM pam2;
void solve() {
    ll ans = 0;
    string s;
    cin >> n >> s;
    pam1.work(s);
    pam2.work(s + s);
    auto &t1 = pam1.t, t2 = pam2.t;
    vector<bool> mp(pam2.tot + 5, 0);
    for (int i = 0; i < n; i++) {
        int u1 = pam1.idpos[i], u2 = pam2.idpos[i];
        if (mp[u2])
            continue;
        mp[u2] = true;
        int cnt = t2[u2].cnt - t1[u1].cnt;
        ans += (((ll)cnt * cnt % mod) * (ll)t2[u2].len) % mod;
        ans %= mod;
    }
    for (int i = n; i < 2 * n; i++) {
        int u = pam2.idpos[i];
        if (mp[u])
            continue;
        mp[u] = true;
        if (t2[u].len > n)
            continue;
        int cnt = t2[u].cnt;
        ans += (((ll)cnt * cnt % mod) *(ll) t2[u].len) % mod;
        ans %= mod;
    }
    cout << ans << endl;
}
signed main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
#ifdef LOCAL
    double starttime = clock();
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
#endif
    int t = 1;
    // cin >> t;
    while (t--) solve();
#ifdef LOCAL
    double endtime = clock();
    cerr << "Time Used: " << (double)(endtime - starttime) / CLOCKS_PER_SEC * 1000 << " ms" << endl;
#endif
    return 0;
}