#pragma GCC optimize("Ofast")
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
#include <cstdio>
#include <ctime>
#include <assert.h>
#include <chrono>
#include <random>
#include <numeric>
#include <set>
#include <deque>
#include <stack>
#include <sstream>
#include <utility>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <tuple>
#include <array>
#include <bitset>
using namespace std;
typedef long long int ll;
typedef unsigned long long ull;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) {
    return (ull)rng() % B;
}
inline double time() {
    return static_cast<double>(chrono::duration_cast<chrono::nanoseconds>(chrono::steady_clock::now().time_since_epoch()).count()) * 1e-9;
}

constexpr ll mod = 998244353;

ll mod_pow(ll a,ll b){
    a%=mod;
    if(b==0)return 1;
    if(b==1)return a;
    ll res=mod_pow(a,b/2)%mod;
    res*=res; res%=mod;
    if(b%2)res*=a;
    return res%mod;
}

struct perm{
private:
    int sz;
public:
    vector<ll> p,invp;
    perm(int n){
        sz=n+1;
        p.resize(sz),invp.resize(sz);
        p[0]=1;
        for(int i=1;i<=sz-1;i++){
            p[i]=p[i-1]*i%mod;
        }
        invp[sz-1]=mod_pow(p[sz-1],mod-2);
        for(int i=sz-2;i>=0;i--){
            invp[i]=invp[i+1]*(i+1)%mod;
        }
    }
    ll comb(ll x,ll y){
        if(x<y||y<0)return 0;
        return (p[x]*invp[x-y]%mod)*invp[y]%mod;
    }
};
perm p(1<<20);

const int N = 5001;

ll dp[27][N];
// dp[i][j]:i文字見て総和jの総数

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    string s; cin >> s;
    int n = s.size();
    vector<int> v(26);
    for(char c:s){
        v[c-'a']++;
    }
    dp[0][0] = 1;
    for(int i=0;i<26;i++){
        for(int j=0;j<=n;j++){
            if(dp[i][j] == 0)continue;
            ll ad = dp[i][j];
            for(int k=0;k<=v[i];k++){
                if(k){
                    ad *= (j+k);
                    ad %= mod;
                }
                dp[i+1][j+k] += ad*p.invp[k]%mod;
                if(dp[i+1][j+k] >= mod){
                    dp[i+1][j+k] -= mod;
                }
            }
        }
    }
    ll res = 0;
    for(int i=0;i<=n;i++){
        res += dp[26][i];
    }
    res--;
    cout << res%mod << endl;
}