#include <bits/stdc++.h>
using namespace std;
char Buff[(1 << 18)], *_S = Buff, *_T = Buff;
template <typename T>
T read(void) {
  T f = 1, num = 0;
  char c = (_S == _T && (_T = (_S = Buff) + fread(Buff, 1, (1 << 18), stdin),
                         _S == _T)
                ? EOF
                : *_S++);
  while (!isdigit(c)) {
    if (c == '-') f = -f;
    c = (_S == _T &&
                 (_T = (_S = Buff) + fread(Buff, 1, (1 << 18), stdin), _S == _T)
             ? EOF
             : *_S++);
  }
  while (isdigit(c)) {
    num = (num << 3) + (num << 1) + (c ^ 48);
    c = (_S == _T &&
                 (_T = (_S = Buff) + fread(Buff, 1, (1 << 18), stdin), _S == _T)
             ? EOF
             : *_S++);
  }
  return f * num;
}
long long fac[5105], inv[5105];
long long power(long long a, long long x) {
  long long ans = 1;
  while (x) {
    if (x & 1) ans = ans * a % 998244353;
    a = a * a % 998244353;
    x >>= 1;
  }
  return ans;
}
inline long long C(int n, int m) {
  return n < m || n < 0 || m < 0
             ? 0
             : fac[n] * inv[m] % 998244353 * inv[n - m] % 998244353;
}
long long solve(int n, int m, int lim) {
  if (n == 0) return 1;
  long long sum = 0;
  for (register int i = 0; i <= m; i++)
    sum = (sum + ((i & 1) ? -1 : 1) * C(m, i) % 998244353 *
                     C(n - i * lim + m - 1, m - 1) % 998244353) %
          998244353;
  return (sum % 998244353 + 998244353) % 998244353;
}
int main() {
  fac[0] = inv[0] = 1;
  for (register int i = 1; i < 5105; i++) fac[i] = fac[i - 1] * i % 998244353;
  inv[5105 - 1] = power(fac[5105 - 1], 998244353 - 2);
  for (register int i = 5105 - 2; i; i--)
    inv[i] = inv[i + 1] * (i + 1) % 998244353;
  int p = read<int>(), s = read<int>(), r = read<int>();
  long long answer = 0;
  for (register int x = r; x <= s; x++)
    for (register int i = 1; i <= p; i++)
      if (i * x + (p - i) * (x - 1) >= s && i * x <= s)
        answer = (answer + solve(s - i * x, p - i, x) * C(p - 1, i - 1) %
                               998244353 * power(i, 998244353 - 2)) %
                 998244353;
  printf("%lld\n",
         answer * power(C(s - r + p - 1, p - 1), 998244353 - 2) % 998244353);
  return 0;
}
