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;
template <class T>
inline void rd(T &x) {
  char ch;
  x = 0;
  bool fl = false;
  while (!isdigit(ch = getchar())) (ch == '-') && (fl = true);
  for (x = (ch ^ '0'); isdigit(ch = getchar()); x = x * 10 + (ch ^ '0'))
    ;
  (fl == true) && (x = -x);
}
template <class T>
inline void output(T x) {
  if (x / 10) output(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void ot(T x) {
  if (x < 0) putchar('-'), x = -x;
  output(x);
  putchar(' ');
}
template <class T>
inline void prt(T a[], int st, int nd) {
  for (register int i = st; i <= nd; ++i) ot(a[i]);
  putchar('\n');
}
namespace Miracle {
const int N = 4008;
const int mod = 998244353;
int n, k, l;
int qm(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = (long long)ret * x % mod;
    x = (long long)x * x % mod;
    y >>= 1;
  }
  return ret;
}
int ad(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
int f[N][N][2];
int main() {
  rd(n);
  rd(k);
  rd(l);
  f[0][0][0] = 1;
  for (register int i = 1; i <= 2 * n + 1; ++i) {
    for (register int j = 0; j <= i; ++j) {
      for (register int x = 0; x <= 1; ++x) {
        if (i + j + (1 - x) <= 2 * n + 1) {
          f[i][j][x] =
              ad(f[i][j][x], (long long)f[i - 1][j + 1][x] * (j + 1) % mod);
          if (j > 0) f[i][j][x] = ad(f[i][j][x], f[i - 1][j - 1][x]);
          if (x == 1 && j >= k) f[i][j][x] = ad(f[i][j][x], f[i - 1][j][0]);
        }
      }
    }
  }
  long long jie = 1;
  for (register int i = n + 1; i <= 2 * n + 1; ++i)
    jie = (long long)jie * i % mod;
  long long ans =
      (long long)f[2 * n + 1][0][1] * qm(2, n) % mod * qm(jie, mod - 2) % mod;
  cout << (long long)ans * l % mod;
  return 0;
}
}  // namespace Miracle
signed main() {
  Miracle::main();
  return 0;
}