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 N = 5005;
const int Md = 998244353;
inline int Add(const int &x, const int &y) {
  return (x + y >= Md) ? (x + y - Md) : (x + y);
}
inline int Sub(const int &x, const int &y) {
  return (x - y < 0) ? (x - y + Md) : (x - y);
}
inline int Mul(const int &x, const int &y) { return (long long)x * y % Md; }
int Powe(int x, int y = Md - 2) {
  int ans = 1;
  for (; y; y >>= 1, x = (long long)x * x % Md)
    if (y & 1) ans = 1ll * ans * x % Md;
  return ans;
}
int n, k, l;
int fac[N], Inv[N], f[N][N];
void Init() {
  fac[0] = Inv[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = Mul(fac[i - 1], i);
  Inv[N - 1] = Powe(fac[N - 1]);
  for (int i = N - 2; i; i--) Inv[i] = Mul(Inv[i + 1], i + 1);
}
int main() {
  scanf("%d%d%d", &n, &k, &l);
  Init();
  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(f[i][j], j));
    }
  }
  int ans = 0;
  int len = Mul(l, Powe((n << 1) + 1));
  for (int i = 1; i <= (n << 1) - 1; i++) {
    for (int j = k; j <= min(n, i); j++)
      ans = Add(ans, Mul(f[i][j], Mul(f[(n << 1) - i][j], fac[j])));
  }
  ans = Mul(ans, Powe(f[n << 1][0]));
  ans = Mul(ans, len);
  printf("%d\n", ans);
  return 0;
}