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;
inline int gi() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum;
}
const int maxn = 4005, mod = 998244353;
int n, m, k, l, ans;
int f[maxn][maxn][2], inv[maxn];
int main() {
  n = gi();
  k = gi();
  l = gi();
  m = 2 * n + 1;
  f[0][0][0] = 1;
  for (int i = 1; i <= m; ++i)
    for (int j = 0; j < i; ++j) {
      f[i][j + 1][0] = (f[i][j + 1][0] + f[i - 1][j][0]) % mod;
      f[i][j + 1][1] = (f[i][j + 1][1] + f[i - 1][j][1]) % mod;
      if (j)
        f[i][j - 1][0] = (f[i][j - 1][0] + (long long)f[i - 1][j][0] * j) % mod;
      if (j)
        f[i][j - 1][1] = (f[i][j - 1][1] + (long long)f[i - 1][j][1] * j) % mod;
      if (j >= k) f[i][j][1] = (f[i][j][1] + f[i - 1][j][0]) % mod;
    }
  ans = (long long)l * f[m][0][1] % mod;
  for (int i = 1; i <= n; ++i) ans = 2ll * ans * i % mod;
  inv[1] = 1;
  for (int i = 2; i <= m; ++i)
    inv[i] = (long long)inv[mod % i] * (mod - mod / i) % mod,
    ans = (long long)ans * inv[i] % mod;
  printf("%d\n", ans);
  return 0;
}