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;
int const p = 998244353;
int fac[4005], inv[4005], iv[4005];
int c(int n, int m) { return 1ll * fac[n] * inv[m] % p * inv[n - m] % p; }
int gamma(int n, int m) {
  return 1ll * fac[n - 1] * fac[m - 1] % p * inv[n + m - 1] % p;
}
int main() {
  int n, k, l, ans = 0, now = 1;
  cin >> n >> k >> l;
  fac[0] = fac[1] = inv[0] = inv[1] = iv[1] = 1;
  for (int i = 2; i <= 2 * n + 1; i++)
    iv[i] = p - 1ll * iv[p % i] * (p / i) % p;
  for (int i = 2; i <= 2 * n + 1; i++)
    fac[i] = 1ll * fac[i - 1] * i % p, inv[i] = 1ll * inv[i - 1] * iv[i] % p;
  for (int i = 1; i <= k; i++) now = 2 * now % p;
  for (int i = k; i <= n; i++, now = 2 * now % p)
    ans = (ans + 1ll * (((i - k) & 1) ? p - 1 : 1) * c(i - 1, i - k) % p *
                     c(n, i) % p * now % p * gamma(i + 1, i + 1)) %
          p;
  cout << 1ll * l * ans % p;
  return 0;
}