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;
long long n, k, l, f[2050][2050][2];
long long powM(long long a, long long t) {
  long long ans = 1;
  while (t) {
    if (t & 1) ans = ans * a % 998244353;
    a = a * a % 998244353;
    t >>= 1;
  }
  return ans;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &k, &l);
  f[0][0][0] = 1;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= i; j++) {
      if (i) {
        f[i][j][0] = (f[i][j][0] + f[i - 1][j][0]) % 998244353;
        f[i][j][1] = (f[i][j][1] + f[i - 1][j][1]) % 998244353;
      }
      if (j) {
        f[i][j][0] =
            (f[i][j][0] + 1ll * (i - j + 1) * f[i][j - 1][0]) % 998244353;
        f[i][j][1] =
            (f[i][j][1] + 1ll * (i - j + 1) * f[i][j - 1][1]) % 998244353;
      }
      if (i - j >= k) f[i][j][1] = (f[i][j][1] + f[i][j][0]) % 998244353;
    }
  long ans = f[n][n][1];
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= i; j++) f[i][j][0] = f[i][j][1] = 0;
  f[0][0][0] = 1;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= i; j++) {
      if (i) {
        f[i][j][0] = (f[i][j][0] + f[i - 1][j][0]) % 998244353;
        f[i][j][1] = (f[i][j][1] + f[i - 1][j][1]) % 998244353;
      }
      if (j) {
        f[i][j][0] =
            (f[i][j][0] + 1ll * (i - j + 1) * f[i][j - 1][0]) % 998244353;
        f[i][j][1] =
            (f[i][j][1] + 1ll * (i - j + 1) * f[i][j - 1][1]) % 998244353;
      }
      f[i][j][1] = (f[i][j][1] + f[i][j][0]) % 998244353;
    }
  printf("%I64d",
         l * ans % 998244353 * powM(f[n][n][1], 998244353 - 2) % 998244353);
  return 0;
}