#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const ll INFLL = 1LL<<60;
const ll MOD = 998244353;
const double INFD = 1.0E10;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, -1, 0, 1};
//const int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
//const int dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
using Pair = pair<ll, ll>;
using Graph = vector<vector<ll, ll> >;
using mint = atcoder::modint998244353;


struct combination{
    vector<mint> fact, ifact;
    combination(int n):fact(n+1), ifact(n+1){
        assert(n < MOD);
        fact[0] = 1;
        for (int i=1; i <= n; i++) fact[i] = fact[i-1] * i;
        ifact[n] = fact[n].inv();
        for (int i=n; i >= 1; i--) ifact[i-1] = ifact[i] * i;
    }
    
    mint operator()(int n, int k){
        if (k < 0 || n < k) return 0;
        return fact[n] * ifact[k] * ifact[n-k];
    }
} comb(2000005);


int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    int n; string s; cin >> n >> s;
    if (s[0] != s[1]) cout << mint(2).pow(n / 2).val() << endl;
    else{
        mint ans = 0;
        for (int i = 0; i <= n / 2; i++){
            ans += comb(n / 2, i) * comb(n / 2, i);
        }
        cout << ans.val() << endl;
    }

    return 0;
}