Problem: Japate, while traveling through the forest of Mala, saw N bags of gold lying in a row. Each bag has some distinct weight of gold between 1 to N. Japate can carry only one bag of gold with him, so he uses the following strategy to choose a bag.

Initially, he starts with an empty bag (zero weight). He considers the bags in some order. If the current bag has a higher weight than the bag in his hand, he picks the current bag.

Japate put the bags in some order. Japate realizes that he will pick A bags, if he starts picking bags from the front, and will pick B bags, if he starts picking bags from the back. By picking we mean replacing the bag in his hand with the current one.

Now he wonders how many permutations of bags are possible, in which he picks A bags from the front and B bags from back using the above strategy.

Since the answer can be very large, output it modulo 998244353.

Input

The only line of input contains three space separated integers N (1 ≤ N ≤ 105), A and B (0 ≤ A, B ≤ N).

Output

Output a single integer — the number of valid permutations modulo 998244353.

Examples

Input

1 1 1


Output

1

Input

2 1 1


Output

0

Input

2 2 1


Output

1

Input

5 2 2


Output

22

Note

In sample case 1, the only possible permutation is [1]

In sample cases 2 and 3, only two permutations of size 2 are possible:{[1, 2], [2, 1]}. The values of a and b for first permutation is 2 and 1, and for the second permutation these values are 1 and 2. 

In sample case 4, out of 120 permutations of [1, 2, 3, 4, 5] possible, only 22 satisfy the given constraints of a and b.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-9;
const int maxn = 131072;
namespace NTT {
const int mod = 998244353;
int qp(long long base, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) (res *= base) %= mod;
    (base *= base) %= mod;
    n >>= 1;
  }
  return (int)res;
}
int rev[maxn << 1];
void change(vector<int> &x, int len) {
  int bit = 0;
  while ((1 << bit) < len) bit++;
  for (int i = (1); i <= (int)(len - 1); ++i) {
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    if (i < rev[i]) swap(x[i], x[rev[i]]);
  }
}
void ntt(vector<int> &x, int lim, int opt) {
  change(x, lim);
  for (int m = 2; m <= lim; m <<= 1) {
    int k = m >> 1;
    int gn = qp(3, (mod - 1) / m);
    for (int i = 0; i < lim; i += m) {
      int g = 1;
      for (int j = 0; j < k; ++j, g = 1ll * g * gn % mod) {
        int tmp = 1ll * x[i + j + k] * g % mod;
        x[i + j + k] = (x[i + j] - tmp + mod) % mod;
        x[i + j] = (x[i + j] + tmp) % mod;
      }
    }
  }
  if (opt == -1) {
    reverse(x.begin() + 1, x.begin() + lim);
    int inv = qp(lim, mod - 2);
    for (int i = (0); i <= (int)(lim - 1); ++i) x[i] = 1ll * x[i] * inv % mod;
  }
}
}  // namespace NTT
using NTT::ntt;
long long qp(long long base, long long n) {
  base %= mod;
  long long res = 1;
  while (n) {
    if (n & 1) (res *= base) %= mod;
    (base *= base) %= mod;
    n >>= 1;
  }
  return res;
}
void PolyInv(int deg, vector<int> &f, vector<int> &g) {
  if (deg == 1) {
    g[0] = qp(f[0], mod - 2);
    return;
  }
  PolyInv((deg + 1) >> 1, f, g);
  int lim = 1;
  while (lim < (deg << 1)) lim <<= 1;
  vector<int> h(lim);
  for (int i = (0); i <= (int)(deg - 1); ++i) h[i] = f[i];
  ntt(g, lim, 1), ntt(h, lim, 1);
  for (int i = (0); i <= (int)(lim - 1); ++i)
    g[i] = 1LL * (2LL + mod - 1LL * g[i] * h[i] % mod) % mod * g[i] % mod;
  ntt(g, lim, -1);
  for (int i = (deg); i <= (int)(lim - 1); ++i) g[i] = 0;
}
namespace Comb {
long long Finv[maxn + 10], fac[maxn + 10], inv[maxn + 10];
long long qp(long long base, long long n) {
  long long res = 1;
  base %= mod;
  while (n) {
    if (n & 1) (res *= base) %= mod;
    (base *= base) %= mod;
    n >>= 1;
  }
  return res;
}
void init() {
  int n = maxn;
  inv[1] = 1;
  for (int i = 2; i <= n; ++i) inv[i] = ((mod - mod / i) * inv[mod % i]) % mod;
  fac[0] = Finv[0] = 1;
  for (int i = 1; i <= n; ++i)
    fac[i] = fac[i - 1] * i % mod, Finv[i] = Finv[i - 1] * inv[i] % mod;
}
long long C(long long n, long long m) {
  if (m < 0 || m > n) return 0;
  return fac[n] * Finv[n - m] % mod * Finv[m] % mod;
}
}  // namespace Comb
using Comb::fac;
using Comb::Finv;
using Comb::inv;
void derivative(int deg, vector<int> &f, vector<int> &g) {
  for (int i = (0); i <= (int)(deg - 2); ++i)
    g[i] = 1LL * f[i + 1] * (i + 1) % mod;
  g[deg - 1] = 0;
}
void integrate(int deg, vector<int> &f, vector<int> &g) {
  for (int i = (deg - 1); i >= (int)(1); --i)
    g[i] = 1LL * f[i - 1] * inv[i] % mod;
  g[0] = 0;
}
void PolyLn(int deg, vector<int> &f, vector<int> &g) {
  assert(f[0] == 1);
  int lim = 1;
  while (lim < (deg << 1)) lim <<= 1;
  vector<int> f_(lim), _f(lim);
  derivative(deg, f, f_);
  PolyInv(deg, f, _f);
  ntt(_f, lim, 1), ntt(f_, lim, 1);
  for (int i = (0); i <= (int)(lim - 1); ++i) f_[i] = 1LL * f_[i] * _f[i] % mod;
  ntt(f_, lim, -1);
  integrate(deg, f_, g);
}
void PolyExp(int deg, vector<int> &f, vector<int> &g) {
  assert(f[0] == 0);
  if (deg == 1) {
    g[0] = 1;
    return;
  }
  PolyExp((deg + 1) >> 1, f, g);
  int lim = 1;
  while (lim < (deg << 1)) lim <<= 1;
  vector<int> gLn(lim);
  PolyLn(deg, g, gLn);
  for (int i = (0); i <= (int)(deg - 1); ++i)
    gLn[i] = (1LL * f[i] - gLn[i] + mod) % mod;
  gLn[0] = (gLn[0] + 1) % mod;
  ntt(g, lim, 1), ntt(gLn, lim, 1);
  for (int i = (0); i <= (int)(lim - 1); ++i) g[i] = 1LL * g[i] * gLn[i] % mod;
  ntt(g, lim, -1);
  for (int i = (deg); i <= (int)(lim - 1); ++i) g[i] = 0;
}
void Polyex1(int deg, vector<int> &f, int k, vector<int> &g) {
  vector<int> h(deg);
  PolyLn(deg, f, h);
  for (int i = (0); i <= (int)(deg - 1); ++i) h[i] = 1LL * h[i] * k % mod;
  PolyExp(deg, h, g);
}
void Polyex2(int deg, vector<int> &f, int k, vector<int> &g) {
  vector<int> h(deg), f_(deg);
  int id = 0, f_id, inv, base;
  while (!f[id]) id++;
  f_id = f[id];
  base = qp(f_id, k), inv = qp(f_id, mod - 2);
  for (int i = (0); i <= (int)(deg - id - 1); ++i)
    f_[i] = 1LL * f[i + id] * inv % mod;
  PolyLn(deg, f_, h);
  for (int i = (0); i <= (int)(deg - 1); ++i) h[i] = 1LL * h[i] * k % mod;
  PolyExp(deg, h, g);
  for (int i = (deg - 1); i >= (int)(k * id); --i)
    g[i] = 1LL * g[i - k * id] * base % mod;
  for (int i = (k * id - 1); i >= (int)(0); --i) g[i] = 0;
}
void Polyex(int deg, vector<int> &f, int k, vector<int> &g) {
  if (f[0] == 1)
    Polyex1(deg, f, k, g);
  else
    Polyex2(deg, f, k, g);
}
vector<int> g(maxn << 1), f(maxn << 1);
namespace fastIO {
static char buf[100000], *h = buf, *d = buf;
template <typename T>
inline void read(T &x) {
  int f = 1;
  x = 0;
  register char c(
      h == d && (d = (h = buf) + fread(buf, 1, 100000, stdin), h == d) ? EOF
                                                                       : *h++);
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = h == d && (d = (h = buf) + fread(buf, 1, 100000, stdin), h == d) ? EOF
                                                                         : *h++;
  }
  while (c <= '9' && c >= '0')
    x = (x << 1) + (x << 3) + (c ^ 48),
    c = h == d && (d = (h = buf) + fread(buf, 1, 100000, stdin), h == d) ? EOF
                                                                         : *h++;
  x *= f;
}
template <typename T>
void output(T x) {
  if (x < 0) {
    putchar('-');
    x = ~(x - 1);
  }
  static int s[20], top = 0;
  while (x) {
    s[++top] = x % 10;
    x /= 10;
  }
  if (!top) s[++top] = 0;
  while (top) putchar(s[top--] + '0');
}
}  // namespace fastIO
using fastIO::output;
using fastIO::read;
int A[maxn], B[maxn];
int main() {
  Comb::init();
  int n, a, b;
  read(n), read(a), read(b);
  if (a == 0 || b == 0) return printf("0\n"), 0;
  for (int i = (1); i <= (int)(n); ++i) g[i] = inv[i];
  Polyex(n + 1, g, a - 1, f);
  for (int i = (0); i <= (int)(n); ++i)
    A[i] = 1LL * f[i] * Finv[a - 1] % mod * fac[i] % mod;
  for (int i = (0); i <= (int)((maxn << 1) - 1); ++i) f[i] = g[i] = 0;
  for (int i = (1); i <= (int)(n); ++i) g[i] = inv[i];
  Polyex(n + 1, g, b - 1, f);
  for (int i = (0); i <= (int)(n); ++i)
    B[i] = 1LL * f[i] * Finv[b - 1] % mod * fac[i] % mod;
  int ans = 0;
  for (int i = (0); i <= (int)(n - 1); ++i)
    (ans += 1LL * A[i] * B[n - 1 - i] % mod * Comb::C(n - 1, i) % mod) %= mod;
  output(ans), putchar('\n');
  return ~~(0 - 0);
}