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>
const int N = 2005, M = N << 1, ZZQ = 998244353;
int n, k, l, f[M][N], g[M][N], ans;
int qpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % ZZQ;
    a = 1ll * a * a % ZZQ;
    b >>= 1;
  }
  return res;
}
inline int add(int &a, const int &b) {
  a += b;
  if (a >= ZZQ) a -= ZZQ;
}
int main() {
  std::cin >> n >> k >> l;
  f[0][0] = 1;
  for (int i = 1; i <= (n << 1); i++)
    for (int j = 0; j <= n; j++) {
      if (j) {
        add(f[i][j], f[i - 1][j - 1]);
        add(g[i][j], g[i - 1][j - 1]);
        if (j - 1 >= k) add(g[i][j], f[i - 1][j - 1]);
      }
      if (j < n) {
        add(f[i][j], 1ll * f[i - 1][j + 1] * (j + 1) % ZZQ);
        add(g[i][j], 1ll * g[i - 1][j + 1] * (j + 1) % ZZQ);
        if (j + 1 >= k) add(g[i][j], 1ll * f[i - 1][j + 1] * (j + 1) % ZZQ);
      }
    }
  std::cout << 1ll * qpow(f[n << 1][0], ZZQ - 2) * g[n << 1][0] % ZZQ *
                   qpow(n << 1 | 1, ZZQ - 2) % ZZQ * l % ZZQ
            << std::endl;
  return 0;
}