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;
const long long mod = 998244353;
inline long long fastpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
inline long long inv(long long x) { return fastpow(x, mod - 2); }
long long fac[4001], p2[2001], ifac[4001], ip2[2001];
inline void Init(long long n) {
  fac[0] = p2[0] = ifac[0] = ip2[0] = 1;
  for (long long i = 1; i <= 2 * n; i++) {
    fac[i] = fac[i - 1] * i % mod;
    ifac[i] = inv(fac[i]);
  }
  for (long long i = 1; i <= n; i++) {
    p2[i] = p2[i - 1] * 2 % mod;
    ip2[i] = inv(p2[i]);
  }
}
signed main() {
  long long n, K, l;
  scanf("%lld %lld %lld", &n, &K, &l);
  Init(n);
  long long ans = 0;
  for (long long k = K; k <= n; k++) {
    long long tot = 0;
    for (long long i = 0; i <= n - k; i++)
      tot = (tot + fac[k + 2 * i] * fac[2 * n - k - 2 * i] % mod *
                       ifac[n - i - k] % mod * ifac[i]) %
            mod;
    ans = (ans + tot * ifac[k] % mod * ip2[n - k]) % mod;
  }
  ans = ans * ifac[2 * n] % mod * fac[n] % mod * p2[n] % mod;
  printf("%lld\n", ans * inv(2 * n + 1) % mod * l % mod);
  return 0;
}