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;
void open(const char *s) {}
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int Minf = 0x7f7f7f7f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const long long MINF = 0x7f7f7f7f7f7f7f7fll;
const int N = 5050;
const int mod = 998244353;
long long f[N][N / 2][2];
long long ksm(long long a, int p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * a % mod;
    a = a * a % mod;
    p >>= 1;
  }
  return ans;
}
int main() {
  cin.tie(0);
  long long n, k, l;
  cin >> n >> k >> l;
  f[0][0][0] = 1;
  for (int i = (1); i <= (n * 2 + 1); ++i)
    for (int j = (0); j <= (i); ++j) {
      f[i][j][1] = (f[i - 1][j + 1][1] * (j + 1) % mod +
                    (j > 0 ? f[i - 1][j - 1][1] : 0)) %
                   mod;
      if (j >= k) f[i][j][1] = (f[i][j][1] + f[i - 1][j][0]) % mod;
      f[i][j][0] = (f[i - 1][j + 1][0] * (j + 1) % mod +
                    (j > 0 ? f[i - 1][j - 1][0] : 0)) %
                   mod;
    }
  long long ans = f[2 * n + 1][0][1];
  for (int i = (1); i <= (n); ++i) ans = ans * 2 * i % mod;
  for (int i = (1); i <= (2 * n + 1); ++i) ans = ans * ksm(i, mod - 2) % mod;
  cout << ans * l % mod << '\n';
}