#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
//using mint = modint1000000007;
//const int mod = 1000000007;
using mint = modint998244353;
const int mod = 998244353;
//const int INF = 1e9;
//const long long LINF = 1e18;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i,l,r)for(int i=(l);i<(r);++i)
#define rrep(i, n) for (int i = (n-1); i >= 0; --i)
#define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i)
#define all(x) (x).begin(),(x).end()
#define allR(x) (x).rbegin(),(x).rend()
#define endl "\n"
#define P pair<int,int>
template<typename A, typename B> inline bool chmax(A & a, const B & b) { if (a < b) { a = b; return true; } return false; }
template<typename A, typename B> inline bool chmin(A & a, const B & b) { if (a > b) { a = b; return true; } return false; }
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) { return com(n, k); }
	mint com(int n, int k) {
		//負の二項係数を考慮する場合にコメントアウトを外す
		//if (n < 0) return com(-n, k) * (k % 2 ? -1 : 1);
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
	mint comsub(long long n, long long k) {
		if (n - k < k) k = n - k;
		assert(k < (int)fact.size());
		mint val = ifact[k];
		rep(i, k) val *= n - i;
		return val;
	}
	mint div(int x) {
		if (x >= (int)fact.size())return mint(x).inv();
		return fact[x - 1] * ifact[x];
	}
	mint inv(int n, int k) {
		//if (n < 0) return inv(-n, k) * (k % 2 ? -1 : 1);
		if (k < 0 || k > n) return 0;
		return ifact[n] * fact[k] * fact[n - k];
	}
	mint p(int n, int k) { return fact[n] * ifact[n - k]; }
}c(2000006);
mint solve(int sz, int x, int k) {
	// 長さsz以下、x種類はすでに使っている、k種類は最低必要
	vector<mint> dp(27);//new col
	mint ret = 0;
	rep(i, 27) {
		mint val = 0;
		if (1 == i + x) {
			val = sz + 1;
		}
		else {
			val = (mint)(pow_mod(i + x, sz + 1, mod) - 1);
			val /= i + x - 1;
		}
		dp[i] = val;
		rep(j, i)dp[i] -= dp[j] * c(i, j);
	}
	rep(i, 27) {
		if (i + x > 26)continue;
		if (i + x < k)continue;
		ret += dp[i] * c(26 - x, i);
	}
	return ret;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, m, k; cin >> n >> m >> k;
	string s; cin >> s;
	vector<bool>chk(26);
	int cnt = 0;
	mint ans = 0;
	rep(i, s.size()) {
		if (cnt >= k)ans++;
		int x = s[i] - 'a';
		int c0 = 0, c1 = 0;
		rep(j, x) {
			if (!chk[j])c1++;
			else c0++;
		}
		int sz = m - (i + 1);
		ans += solve(sz, cnt, k)*c0;
		ans += solve(sz, cnt + 1, k)*c1;
		if (!chk[s[i] - 'a']) {
			chk[s[i] - 'a'] = true;
			cnt++;
		}
	}
	cout << ans.val() << endl;
	return 0;
}