#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void chkmin(T &a, T b) {
  a < b ? a : a = b;
}
template <typename T>
inline void chkmax(T &a, T b) {
  a < b ? a = b : a;
}
namespace fast_IO {
inline int read_int() {
  register int ret = 0, f = 1;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ret = (ret << 1) + (ret << 3) + int(c - 48);
    c = getchar();
  }
  return ret * f;
}
}  // namespace fast_IO
using namespace fast_IO;
int P, S, R;
long long dbinom[5105][105];
inline long long fast_power(long long x, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res *= x, res %= 998244353;
    x *= x, x %= 998244353, k >>= 1;
  }
  return res;
}
inline long long inv(long long x) { return fast_power(x, 998244353 - 2); }
inline void init() {
  P = read_int(), S = read_int(), R = read_int();
  for (register int i = 0; i <= 5100; i++) dbinom[i][0] = 1;
  for (register int i = 0; i <= 5100; i++)
    for (register int j = 1; j <= i && j <= 100; j++)
      dbinom[i][j] = (dbinom[i - 1][j - 1] + dbinom[i - 1][j]) % 998244353;
}
long long sum1, sum2;
inline long long calc(int n, int m, int x) {
  long long res = 0;
  for (register int i = 0; i <= m && i * x <= n; i++) {
    long long tmp =
        1ll * dbinom[m][i] * dbinom[n - x * i + m - 1][m - 1] % 998244353;
    res += (i & 1) ? -tmp : tmp;
  }
  return (res % 998244353 + 998244353) % 998244353;
}
inline void dp() {
  if (P == 1) {
    puts("1");
    return;
  }
  for (register int x = R; x <= S; x++) {
    if (x * P < S) continue;
    for (register int i = 1; i <= P; i++) {
      if (i * x > S || (P - i) * (x - 1) + i * x < S) continue;
      if (i == P) {
        (sum2 += (i * x == S ? inv(i) : 0)) %= 998244353;
        continue;
      }
      (sum2 += 1ll * dbinom[P - 1][i - 1] * calc(S - i * x, P - i, x) %
               998244353 * inv(i)) %= 998244353;
    }
  }
  sum1 = dbinom[S - R + P - 1][P - 1];
  printf("%lld\n", 1ll * sum2 * inv(sum1) % 998244353);
}
int main() {
  init();
  dp();
  return 0;
}
