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;
void read(int &x) {
  static char c;
  while (!isdigit(c = getchar()))
    ;
  x = c ^ 48;
  while (isdigit(c = getchar())) x = (x * 10) + (c ^ 48);
}
void read(long long &x) {
  static char c;
  while (!isdigit(c = getchar()))
    ;
  x = c ^ 48;
  while (isdigit(c = getchar())) x = (x * 10) + (c ^ 48);
}
void chkmax(int &x, int y) {
  if (y > x) x = y;
}
void chkmin(int &x, int y) {
  if (x < y) x = y;
}
void chkmax(long long &x, long long y) {
  if (y > x) x = y;
}
void chkmin(long long &x, long long y) {
  if (x < y) x = y;
}
const int N = 4005, Mod = 998244353;
void Add(int &x, int y) { (x += y) >= Mod ? x -= Mod : x; }
int Pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % Mod;
    a = 1ll * a * a % Mod, b >>= 1;
  }
  return res;
}
int n, k, l;
int Fac[N + 10], Iac[N + 10];
void Init() {
  Fac[0] = Iac[0] = 1;
  for (int i = 1; i <= N; ++i) Fac[i] = 1ll * Fac[i - 1] * i % Mod;
  Iac[N] = Pow(Fac[N], Mod - 2);
  for (int i = N - 1; i; --i) Iac[i] = 1ll * Iac[i + 1] * (i + 1) % Mod;
}
int Binom(int n, int m) {
  return 1ll * Fac[n] * Iac[m] % Mod * Iac[n - m] % Mod;
}
void Work() {
  Init();
  read(n), read(k), read(l);
  int Ans = 0, _ = Pow(2, k);
  for (int s = k; s <= n; ++s) {
    int res = 1ll * Binom(n, s) * Binom(s - 1, s - k) % Mod * _ % Mod * Fac[s] %
              Mod * Fac[s] % Mod * Iac[2 * s + 1] % Mod;
    if ((s - k) & 1) res = Mod - res;
    Add(Ans, res), Add(_, _);
  }
  printf("%d\n", 1ll * l * Ans % Mod);
}
int main() { Work(); }