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 add(int a) { return a >= 998244353 ? a - 998244353 : a; }
inline int dec(int a) { return a < 0 ? a + 998244353 : a; }
inline void Add(int &a, int b) { ((a += b) >= 998244353) && (a -= 998244353); }
inline void Dec(int &a, int b) { ((a -= b) < 0) && (a += 998244353); }
int ksm(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % 998244353;
    a = 1ll * a * a % 998244353;
    b >>= 1;
  }
  return ans;
}
int n, k, l, f[2005 * 2][2005], g[2005 * 2][2005], nn, sm[2005 * 2];
int main() {
  scanf("%d%d%d", &n, &k, &l);
  nn = n + n;
  g[0][0] = 1;
  f[0][0] = 0;
  for (int i = 1; i <= nn; ++i) {
    for (int j = 0; j <= n; ++j) {
      if (j)
        Add(g[i][j - 1], 1ll * j * g[i - 1][j] % 998244353),
            Add(f[i][j - 1],
                1ll * j * ((j >= k) * g[i - 1][j] % 998244353 + f[i - 1][j]) %
                    998244353);
      Add(g[i][j + 1], g[i - 1][j]);
      Add(f[i][j + 1], ((j >= k) * g[i - 1][j] + f[i - 1][j]));
    }
  }
  sm[0] = 1;
  for (int i = 1; i <= nn + 1; ++i) sm[i] = 1ll * sm[i - 1] * i % 998244353;
  printf("%d", 1ll * f[nn][0] * sm[n] % 998244353 * ksm(2, n) % 998244353 *
                   ksm(sm[nn + 1], 998244353 - 2) % 998244353 * l % 998244353);
  return 0;
}