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 N = 1 << 18;
const int P = 998244353;
inline int add(int a, int b) {
  int r = a + b;
  return r < P ? r : r - P;
}
inline int sub(int a, int b) {
  int r = a - b;
  return r < 0 ? r + P : r;
}
inline int mul(long long a, long long b) { return a * b % P; }
inline int inv(int a) { return a == 1 ? a : mul(inv(P % a), P - P / a); }
inline int qpm(int a, int b) {
  int r = 1;
  do
    if (b & 1) r = mul(r, a);
  while (a = mul(a, a), b >>= 1);
  return r;
}
int msqrt(int n) {
  if (!n) return 0;
  int q = P - 1, s = 0, z = 2;
  q >>= (s = __builtin_ctzll(q));
  if (s == 1) return qpm(n, (P + 1) / 4);
  while (qpm(z, (P - 1) / 2) == 1) ++z;
  int c = qpm(z, q), t = qpm(n, q), r = qpm(n, (q + 1) / 2), m = s;
  while (t % P != 1) {
    int i = 1;
    while (qpm(t, 1ll << i) != 1) ++i;
    int b = qpm(c, 1ll << (m - i - 1));
    r = mul(r, b);
    c = mul(b, b);
    t = mul(t, c);
    m = i;
  }
  return min(r, P - r);
}
int invs[N], fac[N], ifac[N];
int binom(int n, int k) { return mul(fac[n], mul(ifac[n - k], ifac[k])); }
void ginv() {
  invs[1] = 1;
  fac[0] = ifac[0] = 1;
  for (int i = 2; i != N; ++i) invs[i] = mul(invs[P % i], (P - P / i));
  for (int i = 1; i != N; ++i) fac[i] = mul(fac[i - 1], i);
  for (int i = 1; i != N; ++i) ifac[i] = mul(ifac[i - 1], invs[i]);
}
vector<int> padd(const vector<int>& p1, const vector<int>& p2) {
  int n3 = max(p1.size(), p2.size());
  vector<int> pr = p1;
  pr.resize(n3, 0);
  for (int i = 0; i != p2.size(); ++i) pr[i] = add(pr[i], p2[i]);
  return pr;
}
vector<int> psub(const vector<int>& p1, const vector<int>& p2) {
  int n3 = max(p1.size(), p2.size());
  vector<int> pr = p1;
  pr.resize(n3, 0);
  for (int i = 0; i != p2.size(); ++i) pr[i] = sub(pr[i], p2[i]);
  return pr;
}
vector<int> pmul(const vector<int>& p1, int k) {
  int n1 = p1.size();
  vector<int> p2(n1);
  for (int i = 0; i != n1; ++i) p2[i] = mul(k, p1[i]);
  return p2;
}
vector<int> egf(const vector<int>& g) {
  vector<int> r(g.size());
  for (int i = 0; i != g.size(); ++i) r[i] = mul(g[i], ifac[i]);
  return r;
}
vector<int> iegf(const vector<int>& g) {
  vector<int> r(g.size());
  for (int i = 0; i != g.size(); ++i) r[i] = mul(g[i], fac[i]);
  return r;
}
vector<int> deriv(const vector<int>& p1) {
  int n1 = p1.size();
  vector<int> p2(n1 - 1);
  for (int i = 1; i != n1; ++i) p2[i - 1] = mul(i, p1[i]);
  return p2;
}
vector<int> integ(const vector<int>& p1) {
  int n1 = p1.size();
  vector<int> p2(n1 + 1, 0);
  for (int i = 0; i != n1; ++i) p2[i + 1] = mul(p1[i], invs[i + 1]);
  return p2;
}
namespace NTT1 {
using ::inv;
using ::mul;
const int W = 18, S = 1 << W, g = 3;
int w[S + 1], rev[S << 1], *r[W + 1];
void init() {
  for (int s = 0; s <= W && (r[s] = rev + (1 << s), 1); ++s)
    for (int i = 0; i != (1 << s); ++i)
      r[s][i] = (r[s][i >> 1] >> 1) | ((i & 1) << (s - 1));
  w[0] = 1;
  w[1] = qpm(g, (P - 1) / S);
  for (int i = 2; i <= S; ++i) w[i] = mul(w[i - 1], w[1]);
}
int m, s, im;
int init(int n) {
  for (s = 0, m = 1; m < n; m <<= 1, ++s)
    ;
  im = inv(m);
  return m;
}
void ntt(int* p, int t) {
  for (int i = 0; i != m; ++i)
    if (i < r[s][i]) swap(p[i], p[r[s][i]]);
  for (int i = 1, z = 0; i != m; i <<= 1, z++)
    for (int j = 0; j != m; j += (i << 1))
      for (int k = 0, u, v; k != i; k++)
        u = p[j + k],
        v = mul(w[(t ? (i << 1) - k : k) << W - z - 1], p[i + j + k]),
        p[j + k] = add(u, v), p[i + j + k] = sub(u, v);
  if (t)
    for (int i = 0; i != m; ++i) p[i] = mul(p[i], im);
}
int px[S], py[S];
vector<int> pmul(const vector<int>& p1, const vector<int>& p2, int n = 0) {
  int n1 = p1.size(), n2 = p2.size(), n3 = n1 + n2 - 1;
  init(n3);
  copy_n(p1.begin(), n1, px);
  fill(px + n1, px + m, 0);
  copy_n(p2.begin(), n2, py);
  fill(py + n2, py + m, 0);
  ntt(px, 0);
  ntt(py, 0);
  for (int i = 0; i != m; ++i) px[i] = mul(px[i], py[i]);
  ntt(px, 1);
  vector<int> p3(n3);
  copy_n(px, n3, p3.begin());
  if (n && n3 > n) p3.resize(n);
  return p3;
}
vector<int> pinv(const vector<int>& p1) {
  int n1 = p1.size(), n2 = (n1 + 1) >> 1;
  if (n1 == 1)
    return {inv(p1[0])};
  else {
    vector<int> p2 = pinv(vector<int>(p1.begin(), p1.begin() + n2));
    init(n1 << 1);
    copy_n(p1.begin(), n1, px);
    fill(px + n1, px + m, 0);
    copy_n(p2.begin(), n2, py);
    fill(py + n2, py + m, 0);
    ntt(px, 0);
    ntt(py, 0);
    for (int i = 0; i != m; ++i) px[i] = mul(sub(2, mul(px[i], py[i])), py[i]);
    ntt(px, 1);
    vector<int> p3(n1);
    copy_n(px, n1, p3.begin());
    return p3;
  }
}
}  // namespace NTT1
using NTT1::init;
using NTT1::pinv;
using NTT1::pmul;
vector<int> shift(const vector<int>& a, int d) {
  int n = a.size();
  vector<int> b = a, c(n);
  reverse(b.begin(), b.end());
  for (int i = 0; i != n; ++i) {
    b[i] = mul(b[i], fac[n - i - 1]);
    if (!i)
      c[i] = 1;
    else
      c[i] = mul(c[i - 1], mul(d, invs[i]));
  }
  vector<int> r = pmul(b, c, n);
  reverse(r.begin(), r.end());
  return egf(r);
}
pair<vector<int>, vector<int> > div(const vector<int>& p1,
                                    const vector<int>& p2) {
  int n1 = p1.size(), n2 = p2.size(), n3 = n1 - n2 + 1;
  if (n3 <= 0) return {{0}, p1};
  vector<int> p1r = p1, p2r = p2;
  reverse(p1r.begin(), p1r.end());
  reverse(p2r.begin(), p2r.end());
  p1r.resize(n3, 0);
  p2r.resize(n3, 0);
  vector<int> p3 = pmul(p1r, pinv(p2r), n3);
  reverse(p3.begin(), p3.end());
  vector<int> p4 = psub(p1, pmul(p2, p3));
  p4.resize(n2 - 1, 0);
  return {p3, p4};
}
vector<int> sqrt(const vector<int>& p1) {
  int n1 = p1.size(), n2 = (n1 + 1) >> 1;
  if (n1 == 1) return {::msqrt(p1[0])};
  vector<int> p2 = sqrt(vector<int>(p1.begin(), p1.begin() + n2));
  vector<int> p3 = pmul(p2, 2);
  p3.resize(n1);
  p3 = pinv(p3);
  return pmul(padd(pmul(p2, p2, n1), p1), p3, n1);
}
vector<int> log(const vector<int>& p1) {
  return integ(pmul(deriv(p1), pinv(p1), p1.size() - 1));
}
vector<int> exp(const vector<int>& p1) {
  if (p1.size() == 1) return {1};
  vector<int> p2 = exp({p1.begin(), p1.begin() + (p1.size() + 1 >> 1)});
  p2.resize(p1.size(), 0);
  return pmul(p2, padd(psub({1}, log(p2)), p1), p1.size());
}
vector<int> pow(const vector<int>& p1, int k) {
  int n1 = p1.size(), n2 = n1;
  while (n2 && !p1[n1 - n2]) n2--;
  int n3 = max(n1 - 1ll * (n1 - n2) * k, 0ll);
  if (!n2 || !n3) return vector<int>(n1, 0);
  vector<int> p2(p1.begin() + n1 - n2, p1.begin() + n1 - n2 + n3);
  int c = p2[0];
  p2 = pmul(exp(pmul(log(pmul(p2, ::inv(c))), k)), qpm(c, k));
  p2.resize(n1, 0);
  rotate(p2.begin(), p2.begin() + n3, p2.end());
  return p2;
}
vector<int> eul(const vector<int>& p1) {
  int n = p1.size();
  vector<int> p2 = p1;
  for (int j = 2; j != n; ++j)
    for (int i = 1; i * j < n; ++i)
      p2[i * j] = add(p2[i * j], mul(p1[i], invs[j]));
  return exp(p2);
}
namespace EVAL {
vector<int> est[N];
int eval(const vector<int>& p, int x) {
  int r = 0;
  for (int i = (int)p.size() - 1; i >= 0; --i) r = add(p[i], mul(r, x));
  return r;
}
void eval0(const vector<int>& x, int p, int lb, int rb) {
  if (lb + 1 == rb)
    est[p] = {sub(0, x[lb]), 1};
  else {
    int mid = (lb + rb) >> 1;
    eval0(x, p << 1, lb, mid);
    eval0(x, p << 1 | 1, mid, rb);
    est[p] = pmul(est[p << 1], est[p << 1 | 1]);
  }
}
void eval1(const vector<int>& r, const vector<int>& x, int p, int lb, int rb) {
  vector<int> w = div(r, est[p]).second;
  if (lb + 100 >= rb)
    for (int i = lb; i != rb; ++i) est[0][i] = eval(w, x[i]);
  else {
    int mid = (lb + rb) >> 1;
    eval1(w, x, p << 1, lb, mid);
    eval1(w, x, p << 1 | 1, mid, rb);
  }
}
vector<int> eval(const vector<int>& p, const vector<int>& x) {
  eval0(x, 1, 0, x.size());
  est[0].resize(x.size());
  eval1(p, x, 1, 0, x.size());
  return est[0];
}
}  // namespace EVAL
using EVAL::eval;
vector<int> stirling1_col(int n, int m) {
  vector<int> a(n + 1, 0);
  for (int i = 1; i <= n; ++i) a[i] = invs[i];
  return iegf(pmul(pow(a, m), ifac[m]));
}
istream& operator>>(istream& is, vector<int>& v) {
  for (int& i : v) is >> i;
  return is;
}
ostream& operator<<(ostream& os, const vector<int>& v) {
  for (int i : v) os << i << ' ';
  return os;
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  init();
  ginv();
  int n, a, b;
  scanf("%d%d%d", &n, &a, &b);
  int ans = 0;
  if (a && b) {
    vector<int> colA = stirling1_col(n, a - 1);
    vector<int> colB = stirling1_col(n, b - 1);
    for (int i = a; i <= n - b + 1; ++i)
      ans = add(ans, mul(binom(n - 1, i - 1), mul(colA[i - 1], colB[n - i])));
  }
  printf("%d\n", ans);
  return 0;
}