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 maxn = 400100;
const int maxm = 500000 + 10;
const int maxt = 3000000 + 100;
const int maxk = 10 + 3;
const long long unit = 1LL;
const int INF = 0x3f3f3f3f;
const long long Inf = 0x3f3f3f3f3f3f3f3fLL;
const double eps = 1e-3;
const double inf = 1e15;
const long double pi = acos(-1.0);
const long long mod = 998244353, gg = 3, inv2 = 499122177;
inline long long read() {
  long long x = 0;
  bool t = false;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') t = true, ch = getchar();
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
  return t ? -x : x;
}
inline long long add(long long x, long long y) {
  return x + y >= mod ? x + y - mod : x + y;
}
inline long long sub(long long x, long long y) {
  return x < y ? x - y + mod : x - y;
}
inline long long mul(long long x, long long y) {
  return x * y - x * y / mod * mod;
}
inline long long upd(long long a) { return a += a >> 31 & mod; }
struct EasyMath {
  inline long long Quick_Pow(long long a, long long b) {
    long long sum = 1;
    for (; b; b >>= 1, a = (long long)a * a % mod)
      if (b & 1) sum = sum * a % mod;
    return sum;
  }
  inline long long inv(long long x) { return Quick_Pow(x, mod - 2); }
} em;
struct NTT {
  int n, k, rev[maxn];
  long long omega[22][maxn], iomega[22][maxn];
  void calc() {
    for (int k = 0; k < 20; ++k) {
      int n = (1 << k);
      const long long x = em.Quick_Pow(gg, (mod - 1) / n);
      omega[k][0] = iomega[k][0] = 1;
      for (int i = 1; i < n; i++)
        omega[k][i] = omega[k][i - 1] * x % mod,
        iomega[k][i] = em.inv(omega[k][i]);
    }
  }
  void init(int n) {
    this->n = n;
    k = (int)log2(n);
    for (int i = 0; i < n; i++)
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (k - 1));
  }
  void transform(long long *a, const long long *omega) {
    for (int i = 0; i < n; ++i)
      if (i < rev[i]) swap(a[i], a[rev[i]]);
    for (int len = 2; len <= n; len *= 2) {
      int mid = len >> 1;
      for (long long *p = a; p != a + n; p += len)
        for (int i = 0; i < mid; i++) {
          long long t = omega[n / len * i] * p[mid + i] % mod;
          p[mid + i] = sub(p[i], t);
          p[i] = add(p[i], t);
        }
    }
  }
  void dft(long long *a) { transform(a, omega[k]); }
  void idft(long long *a) {
    transform(a, iomega[k]);
    long long x = em.inv(n);
    for (int i = 0; i < n; i++) a[i] = a[i] * x % mod;
  }
} ntt;
struct Poly {
  void inverse(const long long *a, const int n, long long *b) {
    if (n == 1) {
      b[0] = em.inv(a[0]);
      return;
    }
    int p = n << 1;
    fill(b, b + p, 0);
    inverse(a, n >> 1, b);
    static long long x[maxn];
    copy(a, a + n, x), fill(x + n, x + p, 0);
    ntt.init(p), ntt.dft(x), ntt.dft(b);
    for (int i = 0; i < p; i++)
      b[i] = b[i] * ((2 - x[i] * b[i] % mod + mod) % mod) % mod;
    ntt.idft(b), fill(b + n, b + p, 0);
  }
  void Ginv(const long long *a, const long long n, long long *b) {
    int p = 1;
    while (p < n) p <<= 1;
    inverse(a, p, b);
    fill(b + n, b + p, 0);
  }
  void Multiply(const long long *a1, const int n1, const long long *a2,
                const int n2, long long *ans) {
    int n = 1;
    while (n < n1 + n2) n <<= 1;
    static long long c1[maxn], c2[maxn];
    copy(a1, a1 + n1, c1), fill(c1 + n1, c1 + n, 0);
    copy(a2, a2 + n2, c2), fill(c2 + n2, c2 + n, 0);
    ntt.init(n);
    ntt.dft(c1);
    ntt.dft(c2);
    for (int i = 0; i < n; i++) c1[i] = c1[i] * c2[i] % mod;
    ntt.idft(c1);
    for (int i = 0; i < n1 + n2 - 1; i++) ans[i] = c1[i];
  }
} poly;
long long n, a, b;
long long f[maxn], fac[maxn], ifac[maxn], inv[maxn];
void LEA(const long long *f, const int n, long long *g, int c) {
  if (n == 0) {
    g[0] = 1;
    return;
  }
  static long long a[maxn], b[maxn];
  for (register int i = (0); i <= (n); ++i) {
    a[i] = mul(f[n - i], fac[n - i]);
  }
  b[0] = 1;
  for (register int i = (1); i <= (n); ++i) {
    b[i] = ((((mul(b[i - 1], c)) * fac[i - 1]) % mod) * ifac[i]) % mod;
  }
  poly.Multiply(a, n + 1, b, n + 1, g);
  reverse(g, g + n + 1);
  for (register int i = (0); i <= (n); ++i) {
    g[i] = mul(g[i], ifac[i]);
  }
  fill(g + n + 1, g + 2 * n + 2, 0);
}
void Gao(int n, long long *f) {
  if (n == 0) {
    f[0] = 1;
    return;
  }
  int mid = (n >> 1), m = (mid << 1);
  Gao(mid, f);
  static long long g[maxn], tmp[maxn];
  LEA(f, mid, g, mid);
  poly.Multiply(f, mid + 1, g, mid + 1, tmp);
  if (n & 1) {
    f[0] = tmp[0] * m % mod;
    for (register int i = (1); i <= (m); ++i) {
      f[i] = add(mul(m, tmp[i]), tmp[i - 1]);
    }
    f[m + 1] = tmp[m];
  } else
    for (register int i = (0); i <= (m); ++i) {
      f[i] = tmp[i];
    }
}
long long xx[maxn], yy[maxn];
void GetInv(int n) {
  inv[1] = fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
  for (register int i = (2); i <= (n); ++i) {
    inv[i] = ((mod - mod / i)) * inv[mod % i] % mod;
    fac[i] = fac[i - 1] * i % mod;
    ifac[i] = ifac[i - 1] * inv[i] % mod;
  }
}
long long C(long long n, long long m) {
  return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
int main() {
  ntt.calc();
  n = read(), a = read(), b = read();
  GetInv(2 * n);
  Gao(n - 1, f);
  printf("%lld\n", (f[a + b - 2] * C(a + b - 2, a - 1)) % mod);
  return 0;
}