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>
const int N = 2010, p = 998244353;
int n, l, k, ans, f[2 * N][N][2];
inline void ad(int &x, int y) { x = (x + y) % p; }
inline int fp(int x, int y, int z = 1) {
  for (; y; y >>= 1, x = 1LL * x * x % p)
    if (y & 1) z = 1LL * z * x % p;
  return z;
}
int main() {
  f[0][0][0] = 1;
  scanf("%d%d%d", &n, &k, &l);
  for (int i = 1; i <= 2 * n + 1; ++i)
    for (int j = 0; j <= n; ++j) {
      if (j >= k) ad(f[i][j][1], f[i - 1][j][0]);
      for (int k = 0; k < 2; ++k) {
        if (j && i + j - k <= 2 * n) ad(f[i][j][k], f[i - 1][j - 1][k]);
        ad(f[i][j][k], 1LL * (j + 1) * f[i - 1][j + 1][k] % p);
      }
    }
  ans = 1LL * l * f[2 * n + 1][0][1] % p;
  for (int i = 1; i <= n; ++i) ans = 2LL * ans * i % p;
  for (int i = 1; i <= 2 * n + 1; ++i) ans = 1LL * ans * fp(i, p - 2) % p;
  printf("%d", ans);
}