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 <typename T>
void chkmin(T &a, T b) {
  a = min(a, b);
}
template <typename T>
void chkmax(T &a, T b) {
  a = max(a, b);
}
inline int fread(void) {
  int ret = 0, op = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') op = -1;
  for (; isdigit(ch); ch = getchar()) ret = (ret << 1) + (ret << 3) - '0' + ch;
  return ret * op;
}
const int MAXN = 5e6 + 10;
const int N = 2e5;
const int MAXM = +10;
const int mod = 998244353;
const int inv2 = ((mod + 1) >> 1);
const int gg = 3;
int fpow(int base, int top) {
  int ret = 1;
  for (; top; top >>= 1) {
    if (top & 1) ret = 1ll * ret * base % mod;
    base = 1ll * base * base % mod;
  }
  return ret;
}
namespace Math {
inline bool check(int x) { return fpow(x, (mod - 1) / 2) == 1; }
inline int random(void) { return 1ll * rand() * rand() % mod; }
int w;
struct Maths {
  int a, b;
  Maths(int _a = 0, int _b = 0) : a(_a), b(_b) {}
  Maths operator+(const Maths you) {
    return Maths((a + you.a) % mod, (b + you.b) % mod);
  }
  Maths operator*(const Maths you) {
    Maths ret;
    ret.a = 1ll * a * you.a % mod + 1ll * b * you.b % mod * w % mod;
    ret.a %= mod;
    ret.b = (1ll * a * you.b % mod + 1ll * b * you.a % mod) % mod;
    return ret;
  }
};
Maths Mpow(Maths base, int top) {
  Maths ret = base;
  top--;
  for (; top; top >>= 1) {
    if (top & 1) ret = ret * base;
    base = base * base;
  }
  return ret;
}
inline int Cipolla(int n) {
  if (!check(n)) return -1;
  int rnd, ret;
  while (1) {
    rnd = random();
    w = 1ll * rnd * rnd % mod;
    w = (w - n + mod) % mod;
    if (!check(w)) break;
  }
  ret = Mpow(Maths(rnd, 1), (mod + 1) / 2).a;
  return min(ret, mod - ret);
}
}  // namespace Math
int n, m, rev[MAXN], mxN, lg2[MAXN], inv[MAXN], Gn[MAXN];
inline void process(int need) {
  for (mxN = 1; mxN < need; mxN <<= 1)
    ;
  lg2[1] = 0;
  for (int i = 2; i <= mxN; ++i) lg2[i] = lg2[i >> 1] + 1;
  for (int i = 0; i < mxN; ++i) {
    rev[i] = rev[i >> 1] >> 1;
    if (i & 1) rev[i] |= (mxN >> 1);
  }
  inv[1] = 1;
  for (int i = 2; i <= mxN; ++i)
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
  for (int len = 1; len < mxN; len <<= 1) {
    int sign = fpow(gg, (mod - 1) / (len << 1));
    Gn[len] = 1;
    for (int i = 1; i < len; ++i)
      Gn[len + i] = 1ll * Gn[len + i - 1] * sign % mod;
  }
}
unsigned long long tv[MAXN];
namespace Poly {
inline void NTT(vector<int> &F, int n, int tp) {
  if (tp == -1) reverse(F.begin() + 1, F.end());
  int tn = lg2[mxN / n];
  for (int i = 0; i < n; ++i) tv[i] = F[i];
  for (int i = 0; i < n; ++i)
    if ((rev[i] >> tn) < i) swap(tv[i], tv[rev[i] >> tn]);
  for (int len = 2; len <= n; len <<= 1) {
    for (int i = 0; i < n; i += len) {
      for (int j = i; j < i + len / 2; ++j) {
        int now = 1ll * Gn[(len >> 1) + j - i] * tv[j + len / 2] % mod;
        tv[j + len / 2] = tv[j] - now + mod;
        tv[j] = tv[j] + now;
      }
    }
  }
  for (int i = 0; i < n; ++i) F[i] = tv[i] % mod;
  if (tp == 1) return;
  int invn = fpow(n, mod - 2);
  for (int i = 0; i < n; ++i) F[i] = 1ll * F[i] * invn % mod;
}
vector<int> fix(vector<int> a, int n) {
  a.resize(n);
  return a;
}
vector<int> operator*(vector<int> a, vector<int> b) {
  if (a.empty() || b.empty()) return vector<int>();
  int used = a.size() + b.size() - 1;
  int lim = (1 << (lg2[used - 1] + 1));
  a.resize(lim);
  b.resize(lim);
  NTT(a, lim, 1);
  NTT(b, lim, 1);
  for (int i = 0; i < lim; ++i) a[i] = 1ll * a[i] * b[i] % mod;
  NTT(a, lim, -1);
  return fix(a, used);
}
vector<int> operator+(vector<int> a, vector<int> b) {
  int lim = max(a.size(), b.size());
  a.resize(lim);
  for (int i = 0; i < lim; ++i) {
    a[i] = a[i] + b[i];
    if (a[i] >= mod) a[i] -= mod;
  }
  return a;
}
vector<int> operator-(vector<int> a, vector<int> b) {
  int lim = max(a.size(), b.size());
  a.resize(lim);
  for (int i = 0; i < lim; ++i) {
    a[i] = a[i] - b[i];
    if (a[i] < 0) a[i] += mod;
  }
  return a;
}
vector<int> operator*(vector<int> a, int x) {
  int lim = a.size();
  for (int i = 0; i < lim; ++i) a[i] = 1ll * a[i] * x % mod;
  return a;
}
vector<int> Inv(vector<int> a) {
  int n = a.size();
  if (n == 1) {
    return vector<int>(1, fpow(a[0], mod - 2));
  }
  vector<int> b = Inv(fix(a, (n + 1) >> 1));
  return fix(b * 2 - a * b * b, n);
}
vector<int> Sqrt(vector<int> a) {
  int n = a.size();
  if (n == 1) {
    return vector<int>(1, Math::Cipolla(a[0]));
  }
  vector<int> b = fix(Sqrt(fix(a, (n + 1) >> 1)), n);
  return fix(((b + a * Inv(b))) * inv2, n);
}
vector<int> Deriv(vector<int> a) {
  int n = a.size();
  for (int i = 0; i < n - 1; ++i) {
    a[i] = 1ll * a[i + 1] * (i + 1) % mod;
  }
  a.pop_back();
  return a;
}
vector<int> Intgal(vector<int> a) {
  int n = a.size();
  a.push_back(0);
  for (int i = n; i >= 1; --i) {
    a[i] = 1ll * a[i - 1] * inv[i] % mod;
  }
  a[0] = 0;
  return a;
}
vector<int> Ln(vector<int> a) {
  int lim = a.size();
  a = fix(Intgal(Deriv(a) * Inv(a)), lim);
  return a;
}
vector<int> Exp(vector<int> a) {
  int n = a.size();
  if (n == 1) return vector<int>(1, 1);
  vector<int> b = fix(Exp(fix(a, (n + 1) >> 1)), n);
  return fix(b - b * (Ln(b) - a), n);
}
vector<int> operator/(vector<int> a, vector<int> b) {
  int n = a.size(), m = b.size();
  reverse(a.begin(), a.end()), reverse(b.begin(), b.end());
  a.resize(n - m + 1), b.resize(n - m + 1);
  a = fix(a * Inv(b), n - m + 1);
  reverse(a.begin(), a.end());
  return a;
}
vector<int> operator%(vector<int> a, vector<int> b) {
  int n = a.size(), m = b.size();
  a = a - (a / b) * b;
  return fix(a, m - 1);
}
vector<int> Shift(vector<int> a) {
  int n = a.size();
  a.push_back(0);
  for (int i = n; i >= 1; --i) a[i] = a[i - 1];
  a[0] = 0;
  a.pop_back();
  return a;
}
};  // namespace Poly
using namespace Poly;
const int MAXB = 150 + 10;
namespace Composit {
vector<int> A1[MAXB], A2[MAXB];
vector<int> compInv(vector<int> f) {
  int n = f.size();
  int siz = sqrt(n) + 1;
  int lim = 1;
  for (; lim < n; lim <<= 1)
    ;
  for (int i = 0; i < n - 1; ++i) f[i] = f[i + 1];
  f.pop_back();
  f = fix(Inv(f), lim);
  A1[0] = vector<int>(1, 1);
  A1[0].resize(lim);
  for (int i = 1; i <= siz; ++i) A1[i] = fix(A1[i - 1] * f, lim);
  A2[0] = vector<int>(1, 1);
  A2[0].resize(lim);
  for (int i = 1; i <= siz; ++i) A2[i] = fix(A2[i - 1] * A1[siz], lim);
  vector<int> ret;
  ret.resize(lim);
  for (int i = 0; i < siz; ++i) {
    for (int j = 0; j < siz && i * siz + j < n; ++j) {
      int tmp = 0;
      for (int k = 0; k <= i * siz + j - 1; ++k) {
        tmp += 1ll * A2[i][k] * A1[j][i * siz + j - 1 - k] % mod;
        tmp %= mod;
      }
      ret[i * siz + j] = 1ll * tmp * fpow(i * siz + j, mod - 2) % mod;
    }
  }
  return fix(ret, n);
}
}  // namespace Composit
inline void init(void) {}
vector<int> F, G, H, Ans;
int fac[MAXN], ifac[MAXN];
void StirlingOne(int len) {
  if (!len) {
    Ans = vector<int>(1, 1);
    return;
  }
  if (len == 1) {
    Ans.resize(2);
    Ans[1] = 1;
    return;
  }
  StirlingOne(len >> 1);
  int nlen = len >> 1;
  F.resize(nlen + 1);
  G.resize(nlen + 1);
  for (int i = 0; i <= nlen; ++i) F[i] = 1ll * Ans[i] * fac[i] % mod;
  int tmp = 1;
  for (int i = 0; i <= nlen; ++i)
    G[i] = 1ll * tmp * ifac[i] % mod, tmp = 1ll * nlen * tmp % mod;
  reverse(F.begin(), F.end());
  H = fix(F * G, nlen + 1);
  for (int i = 0; i <= nlen; ++i) F[i] = 1ll * H[nlen - i] * ifac[i] % mod;
  Ans = fix(F * Ans, len + 1);
  if (len & 1) {
    for (int i = len; i >= 1; --i)
      Ans[i] = (Ans[i - 1] + 1ll * Ans[i] * (len - 1) % mod) % mod;
    Ans[0] = 1ll * Ans[0] * (len - 1) % mod;
    return;
  }
}
int A, B;
int binom(int n, int m) {
  return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
inline void work(void) {
  n = fread();
  process(N << 1);
  A = fread(), B = fread();
  fac[0] = 1;
  for (int i = 1; i <= N; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[N] = fpow(fac[N], mod - 2);
  for (int i = N - 1; i >= 0; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  int ling = 0;
  StirlingOne(n - 1);
  if (A + B - 2 > n - 1) {
    printf("0");
    return;
  }
  int ans = 1ll * Ans[A + B - 2] * binom(A + B - 2, A - 1) % mod;
  printf("%d", ans);
}
int main() {
  init();
  work();
  return 0;
}