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;
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
int MOD = 998244353;
void ADD(long long &x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int SIZE = 4e5 + 10;
long long fac[SIZE], inv[SIZE];
long long mypow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % MOD;
    y >>= 1;
    x = x * x % MOD;
  }
  return res;
}
long long C(int x, int y) {
  if (y < 0 || y > x) return 0;
  return fac[x] * inv[y] % MOD * inv[x - y] % MOD;
}
void build() {
  fac[0] = 1;
  for (int i = (1); i < (SIZE); ++i) fac[i] = fac[i - 1] * i % MOD;
  inv[SIZE - 1] = mypow(fac[SIZE - 1], MOD - 2);
  for (int i = SIZE - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % MOD;
}
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int mod = 998244353;
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mult(int a, int b) { return (ll)a * b % mod; }
int powmod(int a, int pw) {
  int res = 1;
  while (pw) {
    if (pw & 1) res = mult(res, a);
    a = mult(a, a);
    pw >>= 1;
  }
  return res;
}
const int inv2 = (mod + 1) / 2;
namespace FFT {
using ld = int;
using T = int;
const int MAXB = 21;
const int MAXN = 1 << MAXB;
int rev[MAXN];
void fft(T *a, int K, bool inv) {
  const int n = 1 << K;
  rev[0] = 0;
  for (int i = 1; i < n; ++i) {
    rev[i] = (rev[i >> 1] >> 1) ^ ((i & 1) << (K - 1));
  }
  for (int i = 0; i < int(n); ++i) {
    if (i < rev[i]) {
      swap(a[i], a[rev[i]]);
    }
  }
  for (int len = 1; len < n; len *= 2) {
    vector<T> roots(len);
    roots[0] = 1;
    if (len > 1) {
      roots[1] = powmod(3, (mod - 1) / (2 * len));
    }
    for (int i = 2; i < len; ++i) {
      roots[i] = mult(roots[i >> 1], roots[(i + 1) >> 1]);
    }
    for (int i = 0; i < n; i += len * 2) {
      for (int j = 0; j < int(len); ++j) {
        const auto &w = roots[j];
        const auto x = a[i + j];
        const auto y = mult(a[i + len + j], w);
        a[i + j] = x;
        add(a[i + j], y);
        a[i + len + j] = x;
        sub(a[i + len + j], y);
        if (inv) {
          a[i + j] = mult(a[i + j], inv2);
          a[i + len + j] = mult(a[i + j + len], inv2);
        }
      }
    }
  }
  if (inv) {
    reverse(a + 1, a + n);
  }
}
vi mult(const vi &a, const vi &b) {
  int K = 0;
  while ((1 << K) < max(((int)(a).size()), ((int)(b).size()))) {
    ++K;
  }
  ++K;
  vector<T> A(1 << K);
  vector<T> B(1 << K);
  for (int i = 0; i < int(((int)(a).size())); ++i) {
    A[i] = a[i];
  }
  for (int i = 0; i < int(((int)(b).size())); ++i) {
    B[i] = b[i];
  }
  fft(A.data(), K, false);
  fft(B.data(), K, false);
  for (int i = 0; i < int(((int)(A).size())); ++i) {
    A[i] = ::mult(A[i], B[i]);
  }
  fft(A.data(), K, true);
  vi res(((int)(A).size()));
  for (int i = 0; i < int(((int)(A).size())); ++i) {
    res[i] = A[i];
  }
  return res;
}
vi brut_mult(const vi &a, const vi &b) {
  vi c(((int)(a).size()) + ((int)(b).size()));
  for (int i = 0; i < int(((int)(a).size())); ++i) {
    for (int j = 0; j < int(((int)(b).size())); ++j) {
      add(c[i + j], ::mult(a[i], b[j]));
    }
  }
  return c;
}
};  // namespace FFT
vector<int> kerker[SIZE];
void get(int id, int ll, int rr) {
  if (ll == rr) {
    kerker[id] = {ll, 1};
    return;
  }
  int mm = (ll + rr) / 2;
  get(id * 2, ll, mm);
  get(id * 2 + 1, mm + 1, rr);
  kerker[id] = FFT::mult(kerker[id * 2], kerker[id * 2 + 1]);
}
int main() {
  int N, A, B;
  R(N, A, B);
  if (N == 1) {
    if (A == 1 && B == 1)
      W(1);
    else
      W(0);
  } else if (A == 0 || B == 0)
    W(0);
  else if (A == 1 && B == 1)
    W(0);
  else {
    build();
    long long an = C(A + B - 2, A - 1);
    if (N == 2) {
      kerker[1] = {1};
    } else {
      get(1, 1, N - 2);
    }
    an = an * kerker[1][A + B - 3] % MOD;
    W(an);
  }
  return 0;
}