#include #include using namespace std; using namespace atcoder; using mint = modint998244353; //using mint = modint1000000007; using ll = long long; using ld = long double; using pll = pair; using tlll = tuple; constexpr ll INF = 1LL << 60; template bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;} template bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;} ll safemod(ll A, ll M) {return (A % M + M) % M;} ll divfloor(ll A, ll B) {if (B < 0) {return divfloor(-A, -B);} return (A - safemod(A, B)) / B;} ll divceil(ll A, ll B) {if (B < 0) {return divceil(-A, -B);} return divfloor(A + B - 1, B);} #define FINALANS(A) do {cout << (A) << '\n'; exit(0);} while (false) int main() { ll N; string S; cin >> N >> S; vector dp(N + 1, vector(2, vector(2, mint(0)))); dp.at(0).at(0).at(0) = 1; for (ll i = 0; i < N; i++) { for (ll b1 = 0; b1 < 2; b1++) { for (ll b2 = 0; b2 < 2; b2++) { for (char c = 'a'; c <= 'z'; c++) { ll nb1 = b1 + (c == 'a' ? 1 : 0); if (nb1 == 2) continue; ll nb2; if (c < S.at(i)) { nb2 = 1; } else if (c == S.at(i)) { nb2 = b2; } else { if (b2 == 0) continue; nb2 = 1; } dp.at(i + 1).at(nb1).at(nb2) += dp.at(i).at(b1).at(b2); } } } } mint ans = dp.at(N).at(1).at(1); cout << ans.val() << endl; /* for (ll i = 0; i < N + 1; i++) { for (ll b1 = 0; b1 < 2; b1++) { for (ll b2 = 0; b2 < 2; b2++) { cerr << i << " " << b1 << " " << b2 << " " << dp.at(i).at(b1).at(b2).val() << endl; } } } //*/ }