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>
#pragma GCC optimize(2)
using namespace std;
const int N = 4005, P = 998244353;
void add(int &a, int b) {
  a += b;
  a >= P ? a -= P : 0;
}
int mul(int a, int b) { return 1ll * a * b - 1ll * a * b / P * P; }
int f[N][N], fac[N], ans;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
int Pow(int x, int t) {
  int res = 1;
  for (; t; t >>= 1, x = 1ll * x * x % P)
    if (t & 1) res = 1ll * res * x % P;
  return res;
}
int main() {
  int n = read() << 1, k = read(), l = read();
  f[0][0] = 1;
  for (register int i = (0); i <= (n); ++i)
    for (register int j = (0); j <= (min(n >> 1, i)); ++j) {
      add(f[i + 1][j + 1], f[i][j]);
      add(f[i + 1][j - 1], mul(f[i][j], j));
    }
  fac[0] = 1;
  for (register int i = (1); i <= (n); ++i) fac[i] = mul(fac[i - 1], i);
  for (register int i = (1); i <= (n); ++i)
    for (register int j = (k); j <= (min(n >> 1, i)); ++j)
      add(ans, mul(fac[j], mul(f[i][j], f[n - i][j])));
  ans = mul(ans, mul(mul(Pow(n + 1, P - 2), l), Pow(f[n][0], P - 2)));
  printf("%d\n", ans);
  return 0;
}