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 upd(register long long &x, register long long y) {
  x += y;
  if (x >= 998244353) x -= 998244353;
  return;
}
inline long long Qpow(register long long a, register long long p) {
  if (p == 0) return 1;
  register long long t = Qpow(a, p / 2);
  t = t * t % 998244353;
  if (p % 2 == 1) t = t * a % 998244353;
  return t;
}
inline long long inv(register long long a) { return Qpow(a, 998244353 - 2); }
long long f[4005][2005][2];
long long n, k, l;
signed main() {
  cin >> n >> k >> l;
  f[0][0][0] = 1;
  for (register long long i = 0; i <= 2 * n; i++)
    for (register long long j = 0; j <= n; j++) {
      if (f[i][j][0] == 0 && f[i][j][1] == 0) continue;
      if (j < n) upd(f[i + 1][j + 1][0], f[i][j][0]);
      if (j < n) upd(f[i + 1][j + 1][1], f[i][j][1]);
      if (j > 0) upd(f[i + 1][j - 1][0], j * f[i][j][0] % 998244353);
      if (j > 0) upd(f[i + 1][j - 1][1], j * f[i][j][1] % 998244353);
      if (j >= k) upd(f[i + 1][j][1], f[i][j][0]);
    }
  cout << l * f[2 * n + 1][0][1] % 998244353 *
              inv((2 * n + 1) % 998244353 * f[2 * n][0][0] % 998244353) %
              998244353
       << endl;
  return 0;
}