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>
inline long long gi() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
inline int pow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % 998244353;
    x = 1ll * x * x % 998244353;
    y >>= 1;
  }
  return ret;
}
int f[4010][2010], p[2010];
int main() {
  int n = gi(), k = gi(),
      l = 1ll * gi() * pow(n << 1 | 1, 998244353 - 2) % 998244353;
  p[0] = 1;
  for (int i = 1; i <= n; ++i) p[i] = 1ll * i * p[i - 1] % 998244353;
  f[0][0] = 1;
  for (int i = 0; i < n << 1; ++i) {
    for (int j = std::min(n, i); ~j; --j) {
      f[i + 1][j + 1] = (f[i + 1][j + 1] + f[i][j]) % 998244353;
      if (j)
        f[i + 1][j - 1] = (f[i + 1][j - 1] + 1ll * f[i][j] * j) % 998244353;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n << 1; ++i)
    for (int j = k; j <= n; ++j)
      ans = (ans + 1ll * f[i][j] * f[(n << 1) - i][j] % 998244353 * p[j]) %
            998244353;
  printf("%lld\n", 1ll * ans * l % 998244353 *
                       pow(f[n << 1][0], 998244353 - 2) % 998244353);
  return 0;
}