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, P = 998244353;
inline void Add(int& a, const int b) {
  if ((a += b) >= P) a -= P;
}
long long Pow(long long a, long long b, long long s = 1) {
  for (; b; b >>= 1, a = a * a % P)
    if (b & 1) s = s * a % P;
  return s;
}
int n, k, l, f[N * 2][N * 2][2], Fac[N * 2];
int main() {
  scanf("%d%d%d", &n, &k, &l), f[0][0][0] = 1;
  for (register int i = 0; i <= n << 1; ++i)
    for (register int j = 0; j <= i; ++j)
      for (register int o = 0; o <= 1; ++o) {
        Add(f[i + 1][j + 1][o], f[i][j][o]);
        if (!o && j >= k) Add(f[i + 1][j][1], f[i][j][o]);
        if (j) Add(f[i + 1][j - 1][o], (long long)f[i][j][o] * j % P);
      }
  for (register int i = *Fac = 1; i <= n * 2 + 1; ++i)
    Fac[i] = (long long)Fac[i - 1] * i % P;
  return printf("%lld\n", f[n << 1 | 1][0][1] * Pow(2, n) % P * Fac[n] % P *
                              Pow(Fac[n << 1 | 1], P - 2) % P * l % P),
         0;
}