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;
const int mod = 998244353;
const int N = 4005;
const int K = 2005;
int n, k, l;
int fac, ans;
int f[N][K][2];
int fpw(int b, int e = mod - 2) {
  if (!e) return 1;
  int ret = fpw(b, e >> 1);
  ret = 1ll * ret * ret % mod;
  if (e & 1) ret = 1ll * ret * b % mod;
  return ret;
}
int main() {
  scanf("%d%d%d", &n, &k, &l);
  f[0][0][0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++)
    for (int j = 0; j <= n; j++)
      for (int x = 0; x <= 1; x++)
        if (f[i - 1][j][x]) {
          if (j)
            f[i][j - 1][x] = (f[i][j - 1][x] + 1ll * f[i - 1][j][x] * j) % mod;
          if (i + j - 1 < 2 * n + x)
            f[i][j + 1][x] = (f[i][j + 1][x] + f[i - 1][j][x]) % mod;
          if (j >= k && !x) f[i][j][1] = (f[i][j][1] + f[i - 1][j][x]) % mod;
        }
  fac = 1;
  for (int i = n + 1; i <= 2 * n + 1; i++) fac = 1ll * fac * i % mod;
  ans = f[2 * n + 1][0][1];
  ans = 1ll * ans * fpw(2, n) % mod;
  ans = 1ll * ans * fpw(fac) % mod * l % mod;
  printf("%d\n", ans);
  return 0;
}