#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#include<map>
#include<vector>
#include<stack>
#include<iomanip>
#include<queue>
#include<set>
#include<functional>
#include<tuple>
#include<bitset>
#include<cassert>
#include<cstdint>
#include<complex>
#include<random>
using namespace std;
bool printb(bool f) {
	if (f)printf("Yes\n");
	else printf("No\n");
	return f;
}
template<class T>
void prt(T t = "", string sep = "\n") { cout << t << sep; return; }
template<class T>
void printl(vector<T> a, string sep = " ") {
	for (int i = 0; i < a.size(); i++) {
		cout << a[i];
		if (i != a.size() - 1)cout << sep;
	}
	if (sep != "\n")cout << "\n";
	return;
}
bool prt_isfixed = false;
template<class T>
void prt_fix(T t, string sep = "\n") {
	if (!prt_isfixed) {
		cout << fixed << setprecision(15);
		prt_isfixed = true;
	}
	prt(t, sep);
}
#define all(a) a.begin(),a.end()
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using uint = unsigned int;
using llong = long long;
using ullong = unsigned long long;
using pii = pair<int, int>;
using pll = pair<llong, llong>;
using pli = pair<llong, int>;
using pil = pair<int, llong>;
template<typename T> using vec2 = vector<vector<T>>;
template<typename T> inline bool chmin(T& a, T b) { return (a > b) ? (a = b, true) : false; }
template<typename T> inline bool chmax(T& a, T b) { return (a < b) ? (a = b, true) : false; }
bool bitIn(llong a, int b) { return ((a >> b) & 1); }
int bitCnt(llong a) {
	int re = 0;
	while (a > 0) {
		if (a & 1)re++;
		a >>= 1;
	}
	return re;
}
llong powL(llong n, llong i) {
	llong re = 1;
	while (i >= 1) {
		if (i & 1) re *= n;
		n *= n;
		i >>= 1;
	}
	return re;
}

llong powL_M(llong n, llong i, llong mod) {
	llong re = 1;
	while (i >= 1) {
		if (i & 1) {
			re *= n;
			re %= mod;
		}
		n *= n;
		n %= mod;
		i >>= 1;
	}
	return re;
}

int  dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
int dx8[8] = { 0,1,1,1,0,-1,-1,-1 }, dy8[8] = { -1,-1,0,1,1,1,0,-1 };

/*
modintクラス。四則演算と累乗が定義されている。
割り算はmodが素数でない時にも使える。(逆元の存在条件注意)
extGCD(),GCD()を含む
*/
template <class T>
T extGCD(T a, T b, T& x, T& y) {
	if (b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	T gcd = extGCD(b, a % b, y, x);
	y -= a / b * x;
	return gcd;
}

template <class T>
T GCD(T a, T b) {
	T x, y;
	return extGCD(a, b, x, y);
}


static const int mod = 998244353; //問題文に合わせて変更すること
class modint {
public:
	long long x;
	modint(long long x = 0) :x((x% mod + mod) % mod) {}
	modint operator-() const {
		return (-x);
	}
	modint& operator+=(const modint& a) {
		if ((x += a.x) >= mod)x -= mod;
		return *this;
	}
	modint& operator-=(const modint& a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	modint& operator*=(const modint& a) {
		(x *= a.x) %= mod;
		return *this;
	}
	modint operator+(const modint& a) const {
		modint res(*this);
		return res += a;
	}
	modint operator-(const modint& a) const {
		modint res(*this);
		return res -= a;
	}
	modint operator*(const modint& a) const {
		modint res(*this);
		return res *= a;
	}

	modint inv() const {
		long long y, c;
		extGCD(x, (long long)mod, y, c);
		return y;
	}
	modint& operator/=(const modint& a) {
		return (*this) *= a.inv();
	}
	modint operator/ (const modint& a) const {
		modint res(*this);
		return res /= a;
	}
	friend ostream& operator<<(ostream& os, const modint& a) {
		os << a.x;
		return os;
	}
};
//pow(a , n) modint型aのn乗のmodを求める
template<class T>
modint powM(modint a, T n) {
	modint re(1);
	while (n > 0) {
		if (n & 1)re *= a;
		a *= a;
		n >>= 1;
	}
	return re;
}
modint sum_numM(modint l, modint r) {
	return (l + r) * (r - l + 1) / 2;
}


int main() {
	int n, m, k;
	cin >> n >> m >> k;	
	string s;
	cin >> s;
	vector dp(m + 1, vector<modint>(27, 0));
	//dp[i][j]=i文字目まで決めてj種類の文字を使ってて、辞書順でsよりも小さいようなもの

	vector<bool> isUsed(27,false);
	//sのi文字までに使われているか
	int cnt = 0;
	modint re;
	for (int i = 1; i <= m; i++) {
		if (i <= n) {
			for (int j = 0; j < s[i-1]-'a'; j++) {
				if (isUsed[j])dp[i][cnt] += 1;
				else dp[i][cnt + 1] += 1;
			}
			if (cnt >= k)re+=1;
			if (!isUsed[s[i - 1] - 'a']) {
				isUsed[s[i - 1] - 'a'] = true;
				cnt++;
			}
		}
		for (int j = 1; j < 27; j++) {
			dp[i][j] += dp[i - 1][j - 1] * (26 - j + 1);
			dp[i][j] += dp[i - 1][j] * j;
		}
	}
	//for (auto i : dp)printl(i);
	for (int i = 1; i <= m; i++) {
		for (int j = k; j < 27;j++) {
			re += dp[i][j];
		}
	}
	prt(re);
}