Problem: Getting closer and closer to a mathematician, Serval becomes a university student on math major in Japari University. On the Calculus class, his teacher taught him how to calculate the expected length of a random subsegment of a given segment. Then he left a bonus problem as homework, with the award of a garage kit from IOI. The bonus is to extend this problem to the general case as follows.

You are given a segment with length l. We randomly choose n segments by choosing two points (maybe with non-integer coordinates) from the given segment equiprobably and the interval between the two points forms a segment. You are given the number of random segments n, and another integer k. The 2n endpoints of the chosen segments split the segment into (2n+1) intervals. Your task is to calculate the expected total length of those intervals that are covered by at least k segments of the n random segments.

You should find the answer modulo 998244353.

Input

First line contains three space-separated positive integers n, k and l (1≤ k ≤ n ≤ 2000, 1≤ l≤ 10^9).

Output

Output one integer — the expected total length of all the intervals covered by at least k segments of the n random segments modulo 998244353.

Formally, let M = 998244353. It can be shown that the answer can be expressed as an irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output the integer equal to p ⋅ q^{-1} mod M. In other words, output such an integer x that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.

Examples

Input


1 1 1


Output


332748118


Input


6 2 1


Output


760234711


Input


7 5 3


Output


223383352


Input


97 31 9984524


Output


267137618

Note

In the first example, the expected total length is \int_0^1 \int_0^1 |x-y|  dx dy = {1\over 3}, and 3^{-1} modulo 998244353 is 332748118.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using lint = long long;
template <class T = int>
using V = vector<T>;
template <class T = int>
using VV = V<V<T> >;
template <unsigned P>
struct ModInt {
  using M = ModInt;
  unsigned v;
  ModInt() : v(0) {}
  template <class Z>
  ModInt(Z x) : v(x >= 0 ? x % P : (P - -x % P) % P) {}
  constexpr ModInt(unsigned v, int) : v(v) {}
  static constexpr unsigned p() { return P; }
  M operator+() const { return *this; }
  M operator-() const { return {v ? P - v : 0, 0}; }
  explicit operator bool() const noexcept { return v; }
  bool operator!() const noexcept { return !(bool)*this; }
  M operator*(M r) const { return M(*this) *= r; }
  M operator/(M r) const { return M(*this) /= r; }
  M operator+(M r) const { return M(*this) += r; }
  M operator-(M r) const { return M(*this) -= r; }
  bool operator==(M r) const { return v == r.v; }
  bool operator!=(M r) const { return !(*this == r); }
  M& operator*=(M r) {
    v = (uint64_t)v * r.v % P;
    return *this;
  }
  M& operator/=(M r) { return *this *= r.inv(); }
  M& operator+=(M r) {
    if ((v += r.v) >= P) v -= P;
    return *this;
  }
  M& operator-=(M r) {
    if ((v += P - r.v) >= P) v -= P;
    return *this;
  }
  M inv() const {
    int a = v, b = P, x = 1, u = 0;
    while (b) {
      int q = a / b;
      swap(a -= q * b, b);
      swap(x -= q * u, u);
    }
    assert(a == 1);
    return x;
  }
  template <class Z>
  M pow(Z n) const {
    if (n < 0) return pow(-n).inv();
    M res = 1;
    for (M a = *this; n; a *= a, n >>= 1)
      if (n & 1) res *= a;
    return res;
  }
  template <class Z>
  friend M operator*(Z l, M r) {
    return M(l) *= r;
  }
  template <class Z>
  friend M operator/(Z l, M r) {
    return M(l) /= r;
  }
  template <class Z>
  friend M operator+(Z l, M r) {
    return M(l) += r;
  }
  template <class Z>
  friend M operator-(Z l, M r) {
    return M(l) -= r;
  }
  friend ostream& operator<<(ostream& os, M r) { return os << r.v; }
  friend istream& operator>>(istream& is, M& r) {
    lint x;
    is >> x;
    r = x;
    return is;
  }
  template <class Z>
  friend bool operator==(Z l, M r) {
    return M(l) == r;
  }
  template <class Z>
  friend bool operator!=(Z l, M r) {
    return !(l == r);
  }
};
using Mint = ModInt<998244353>;
V<Mint> fact, ifact, inv, powB, dfact;
void init(int n, int B = 2) {
  fact.resize(n + 1);
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fact[i] = i * fact[i - 1];
  }
  ifact.resize(n + 1);
  ifact[n] = 1 / fact[n];
  for (int i = n; i > 0; --i) {
    ifact[i - 1] = i * ifact[i];
  }
  powB.resize(n + 1);
  powB[0] = 1;
  for (int i = 0; i < n; ++i) {
    powB[i + 1] = powB[i] * B;
  }
  dfact.resize(n + 1);
  dfact[0] = dfact[1] = dfact[2] = 1;
  for (int i = 2; i < n; ++i) {
    dfact[i + 1] = i * dfact[i - 1];
  }
}
Mint comb(int n, int r) {
  if (r < 0 or r > n) return 0;
  return fact[n] * ifact[r] * ifact[n - r];
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n, k, l;
  cin >> n >> k >> l;
  init(2 * n);
  Mint res = 0;
  for (int i = 1; i < 2 * n; ++i) {
    for (int x = k; x <= n; ++x)
      if ((x & 1) == (i & 1)) {
        if (x > i or x > 2 * n - i) continue;
        res += comb(i, x) * comb(2 * n - i, x) * fact[x] * dfact[i - x] *
               dfact[2 * n - i - x];
      }
  }
  res /= dfact[2 * n];
  res /= 2 * n + 1;
  res *= l;
  cout << res << '\n';
}