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 p = 998244353;
template <typename T>
inline T read() {
  register T sum = 0;
  register char cc = getchar();
  int sym = 1;
  while (cc != '-' && (cc > '9' || cc < '0')) cc = getchar();
  if (cc == '-') sym = -1, cc = getchar();
  sum = sum * 10 + cc - '0';
  cc = getchar();
  while (cc >= '0' && cc <= '9') sum = sum * 10 + cc - '0', cc = getchar();
  return sym * sum;
}
template <typename T>
inline T read(T& a) {
  a = read<T>();
  return a;
}
template <typename T, typename... Others>
inline void read(T& a, Others&... b) {
  a = read(a);
  read(b...);
}
int n, l, K, ans, fac[4010], f[4010][4010];
int add(int x, int y) { return (x + y) % p; }
int mul(int x, int y) { return 1ll * x * y % p; }
int fpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = mul(res, x);
    x = mul(x, x);
    y >>= 1;
  }
  return res;
}
int main() {
  read(n, K, l);
  l = mul(l, fpow((n << 1) + 1, p - 2));
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mul(fac[i - 1], i);
  f[0][0] = 1;
  for (int i = 0; i < n << 1; i++)
    for (int j = 0; j <= min(n, i); j++) {
      f[i + 1][j + 1] = add(f[i + 1][j + 1], f[i][j]);
      if (j) f[i + 1][j - 1] = add(f[i + 1][j - 1], mul(j, f[i][j]));
    }
  for (int i = 1; i <= n << 1; i++)
    for (int j = K; j <= n; j++)
      ans = add(ans, mul(mul(f[i][j], f[(n << 1) - i][j]), fac[j]));
  printf("%d\n", mul(mul(ans, l), fpow(f[n << 1][0], p - 2)));
  return 0;
}