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;
inline void proc_status() {
  ifstream t("/proc/self/status");
  cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>())
       << endl;
}
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline T read() {
  register T sum(0), fg(1);
  register char ch(getchar());
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fg = -1;
  for (; isdigit(ch); ch = getchar()) sum = sum * 10 - '0' + ch;
  return sum * fg;
}
const int MOD = 998244353;
namespace MATH {
const int N = (int)1e6;
inline int fpm(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = (long long)x * x % MOD)
    if (y & 1) res = (long long)res * x % MOD;
  return res;
}
int fac[N + 5], ifac[N + 5];
inline int C(int n, int m) {
  return n < m ? 0 : (long long)fac[n] * ifac[n - m] % MOD * ifac[m] % MOD;
}
inline void init() {
  fac[0] = 1;
  for (int i = 1; i <= N; ++i) fac[i] = (long long)fac[i - 1] * i % MOD;
  ifac[N] = fpm(fac[N], MOD - 2);
  for (int i = N - 1; i >= 0; --i)
    ifac[i] = (long long)ifac[i + 1] * (i + 1) % MOD;
}
}  // namespace MATH
using MATH::C;
using MATH::fac;
using MATH::fpm;
using MATH::ifac;
const int MAXN = (int)2e3;
int n, k, l;
inline void input() { n = read<int>(), k = read<int>(), l = read<int>(); }
inline void solve() {
  static int f[MAXN + 5], g[MAXN + 5];
  for (int i = 1; i <= n; ++i)
    g[i] = (long long)fpm(2, i) * fac[i] % MOD * fac[i] % MOD * ifac[2 * i] %
           MOD * fpm(2 * i + 1, MOD - 2) % MOD * C(n, i) % MOD;
  for (int i = n; i >= 1; --i) {
    f[i] = 0;
    for (int j = i; j <= n; ++j)
      (f[i] += (long long)((j - i) & 1 ? -1 : +1) * C(j, i) * g[j] % MOD) %=
          MOD;
  }
  int ans = 0;
  for (int i = k; i <= n; ++i) (ans += f[i]) %= MOD;
  printf("%lld\n", 1ll * (ans + MOD) % MOD * l % MOD);
}
int main() {
  MATH::init();
  input();
  solve();
  return 0;
}