#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
template<class T> bool chmax(T &a, const T &b) { if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a, const T &b) { if(a>b){a=b;return 1;}return 0;}
//-------------------------------------------------------

int N;
string S;

ll dp[50505][27][27];
const ll mo=998244353;

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	dp[0][26][26]=1;
	cin>>N;
	cin>>S;
	FOR(i,N) {
		
		if(S[i]=='?') {
			FOR(y,27) FOR(k,26) if(k!=y) {
				FOR(x,27) if(x!=k) dp[i+1][y][k]+=dp[i][x][y];
				dp[i+1][y][k]%=mo;
			}
		}
		else {
			int c=S[i]-'a';
			FOR(x,27) FOR(y,27) if(c!=x&&c!=y) (dp[i+1][y][c]+=dp[i][x][y])%=mo;
		}
	}
	
	ll ret=0;
	FOR(x,27) FOR(y,27) ret+=dp[N][x][y];
	cout<<ret%mo<<endl;
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}