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 std::cerr;
using std::endl;
const int N = 4005, P = 998244353;
inline int fpow(int x, int y) {
  int ret = 1;
  for (; y; y >>= 1, x = 1ll * x * x % P)
    if (y & 1) ret = 1ll * ret * x % P;
  return ret;
}
int n, K, L, f[N][N], match[N], ans;
int main() {
  std::cin >> n >> K >> L;
  f[0][0] = 1;
  n = n * 2;
  for (int i = 1; i <= n + 1; ++i) {
    for (int j = 0; j <= i + 1; ++j) {
      if (j != 0) f[i][j] = (f[i][j] + f[i - 1][j - 1]) % P;
      f[i][j] = (f[i][j] + 1ll * f[i - 1][j + 1] * (j + 1)) % P;
    }
  }
  match[0] = 1;
  for (int i = 1; i <= n + 1; ++i) match[i] = 1ll * match[i - 1] * i % P;
  for (int i = 0; i <= n; ++i) {
    for (int j = K; j <= i + 1; ++j) {
      int tmp = 1ll * f[i][j] * f[n - i][j] % P;
      tmp = 1ll * tmp * match[j] % P;
      ans = (ans + tmp) % P;
    }
  }
  ans = 1ll * ans * fpow(f[n][0], P - 2) % P;
  ans = 1ll * ans * L % P;
  ans = 1ll * ans * fpow(n + 1, P - 2) % P;
  std::cout << ans << std::endl;
  return 0;
}