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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
const int INF = 1.01e9;
const int MOD = 998244353;
const int N = (int)1.01e6;
void print(vector<int> a) {
  for (int x : a) cout << x << " ";
  cout << endl;
}
int fct[N], ofct[N];
void add(int &a, long long b) { a = (a + b) % MOD; }
int getC(int n, int k) {
  if (k < 0 || n < k || n < 0) return 0;
  return 1LL * fct[n] * ofct[k] % MOD * ofct[n - k] % MOD;
}
int bin(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1LL * res * a % MOD;
    a = 1LL * a * a % MOD;
    n >>= 1;
  }
  return res;
}
int inv(int x) { return bin(x, MOD - 2); }
void init() {
  fct[0] = 1;
  for (int i = 1; i < N; i++) fct[i] = 1LL * fct[i - 1] * i % MOD;
  ofct[N - 1] = inv(fct[N - 1]);
  for (int i = N - 2; i >= 0; i--) ofct[i] = 1LL * ofct[i + 1] * (i + 1) % MOD;
}
namespace fft {
const int maxBase = 18;
const int maxN = 1 << maxBase;
struct num {
  double x, y;
  num() {}
  num(double xx, double yy) : x(xx), y(yy) {}
  num(double alp) : x(cos(alp)), y(sin(alp)) {}
};
inline num operator+(num a, num b) { return num(a.x + b.x, a.y + b.y); }
inline num operator-(num a, num b) { return num(a.x - b.x, a.y - b.y); }
inline num operator*(num a, num b) {
  return num(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}
inline num conj(num a) { return num(a.x, -a.y); }
const double PI = acos(-1);
num root[maxN];
int rev[maxN];
bool rootsPrepared = false;
void prepRoots() {
  if (rootsPrepared) return;
  rootsPrepared = true;
  root[1] = num(1, 0);
  for (int k = 1; k < maxBase; ++k) {
    num x(2 * PI / (1LL << (k + 1)));
    for (int i = (1LL << (k - 1)); i < (1LL << (k)); ++i) {
      root[2 * i] = root[i];
      root[2 * i + 1] = root[i] * x;
    }
  }
}
int base, N;
int lastRevN = -1;
void prepRev() {
  if (lastRevN == N) return;
  lastRevN = N;
  for (int i = 0; i < (N); i++)
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (base - 1));
}
void fft(num *a, num *f) {
  for (int i = 0; i < (N); i++) f[i] = a[rev[i]];
  for (int k = 1; k < N; k <<= 1)
    for (int i = 0; i < N; i += 2 * k)
      for (int j = 0; j < (k); j++) {
        num z = f[i + j + k] * root[j + k];
        f[i + j + k] = f[i + j] - z;
        f[i + j] = f[i + j] + z;
      }
}
num a[maxN], b[maxN], f[maxN], g[maxN];
long long A[maxN], B[maxN], C[maxN];
void _multMod(int mod, bool eq) {
  for (int i = 0; i < (N); i++) {
    int x = A[i] % mod;
    a[i] = num(x & ((1LL << (15)) - 1), x >> 15);
  }
  for (int i = 0; i < (N); i++) {
    int x = B[i] % mod;
    b[i] = num(x & ((1LL << (15)) - 1), x >> 15);
  }
  fft(a, f);
  if (!eq) {
    fft(b, g);
  } else {
    for (int i = 0; i < N; i++) g[i] = f[i];
  }
  for (int i = 0; i < (N); i++) {
    int j = (N - i) & (N - 1);
    num a1 = (f[i] + conj(f[j])) * num(0.5, 0);
    num a2 = (f[i] - conj(f[j])) * num(0, -0.5);
    num b1 = (g[i] + conj(g[j])) * num(0.5 / N, 0);
    num b2 = (g[i] - conj(g[j])) * num(0, -0.5 / N);
    a[j] = a1 * b1 + a2 * b2 * num(0, 1);
    b[j] = a1 * b2 + a2 * b1;
  }
  fft(a, f);
  fft(b, g);
  for (int i = 0; i < (N); i++) {
    long long aa = f[i].x + 0.5;
    long long bb = g[i].x + 0.5;
    long long cc = f[i].y + 0.5;
    C[i] = (aa + bb % mod * (1LL << (15)) + cc % mod * (1LL << (30))) % mod;
  }
}
void prepAB(int n1, int n2) {
  base = 1;
  N = 2;
  while (N < n1 + n2) base++, N <<= 1;
  for (int i = n1; i < N; ++i) A[i] = 0;
  for (int i = n2; i < N; ++i) B[i] = 0;
  prepRoots();
  prepRev();
}
void mult(int n1, int n2, int mod, bool eq) {
  prepAB(n1, n2);
  _multMod(mod, eq);
}
}  // namespace fft
int slowSubsolve(int n, int a) {
  if (a == 0) {
    return n == 0;
  }
  vector<int> p(n);
  iota(p.begin(), p.end(), 0);
  int ans = 0;
  do {
    int cur = -1;
    int cnt = 0;
    for (int x : p) {
      if (x > cur) {
        cur = x;
        cnt++;
      }
    }
    ans += cnt == a;
  } while (next_permutation(p.begin(), p.end()));
  return ans;
}
vector<int> mul(const vector<int> &a, const vector<int> &b, int k,
                bool eq = 0) {
  for (int i = 0; i < k; i++) {
    fft::A[i] = a[i];
    fft::B[i] = b[i];
  }
  fft::mult(k, k, MOD, eq);
  vector<int> c(k);
  for (int i = 0; i < k; i++) c[i] = fft::C[i];
  return c;
}
pair<vector<int>, vector<int>> bin(vector<int> a, int n1, int n2, int k) {
  vector<int> res1(k);
  res1[0] = 1;
  auto res2 = res1;
  while (n1 || n2) {
    if (n1 & 1) res1 = mul(res1, a, k);
    n1 >>= 1;
    if (n2 & 1) res2 = mul(res2, a, k);
    n2 >>= 1;
    a = mul(a, a, k, 1);
  }
  return {res1, res2};
}
vector<vector<int>> fastSubsolve(int m1, int m2, int n) {
  vector<int> f(n);
  for (int i = 1; i < n; i++) {
    f[i] = 1LL * ofct[i] * fct[i - 1] % MOD;
    if (i % 2 == 0) f[i] = (MOD - f[i]) % MOD;
  }
  auto ff = bin(f, m1, m2, n);
  vector<vector<int>> res;
  for (int it = 0; it < 2; it++) {
    f = ff.first;
    int o = ofct[m1];
    for (int i = 0; i < n; i++) {
      f[i] = 1LL * f[i] * o % MOD;
    }
    for (int i = 0; i < n; i++) {
      f[i] = 1LL * f[i] * fct[i] % MOD;
    }
    for (int i = 0; i < n; i++) {
      if ((i - m1) % 2 != 0) {
        f[i] = (MOD - f[i]) % MOD;
      }
    }
    res.push_back(f);
    swap(m1, m2);
    swap(ff.first, ff.second);
  }
  return res;
}
int slow(int n, int a, int b) {
  if (a == 0 || b == 0) return 0;
  int ans = 0;
  for (int x = 0; x < n; x++) {
    add(ans, 1LL * getC(n - 1, x) * slowSubsolve(x, a - 1) % MOD *
                 slowSubsolve(n - 1 - x, b - 1));
  }
  return ans;
}
int fast(int n, int a, int b) {
  if (a == 0 || b == 0) return 0;
  auto o = fastSubsolve(a - 1, b - 1, n);
  auto va = o[0];
  auto vb = o[1];
  int ans = 0;
  for (int x = 0; x < n; x++) {
    add(ans, 1LL * getC(n - 1, x) * va[x] % MOD * vb[n - 1 - x]);
  }
  return ans;
}
int main() {
  init();
  if (0) {
    int A = (1 << 15) - 1;
    cout << fast(1e5, A, A) << endl;
    cerr << "time = " << (clock() * 1. / CLOCKS_PER_SEC) << "\n";
    return 0;
  }
  if (0) {
    for (int n = 1; n <= 10; n++) {
      cout << n << endl;
      for (int a = 0; a <= n; a++) {
        for (int b = 0; b <= n; b++) {
          int ans1 = fast(n, a, b);
          int ans2 = slow(n, a, b);
          assert(ans1 == ans2);
        }
      }
    }
    return 0;
  }
  int n, a, b;
  while (cin >> n >> a >> b) {
    cout << fast(n, a, b) << endl;
  }
  return 0;
}