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;
long long n, k, l;
long long fakdp[20000];
long long dizdp[20000];
long long us(long long a, long long b) {
  if (a == 1) return 1;
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans *= a;
      ans %= 998244353;
    }
    b >>= 1;
    a *= a;
    a %= 998244353;
  }
  return ans;
}
long long diz(long long nokta) {
  assert(nokta < 20000);
  if (nokta % 2 != 0) return 0;
  if (dizdp[nokta] != -1) return dizdp[nokta];
  long long ans = 1;
  for (long long i = 3; i < nokta; i += 2) {
    ans *= i;
    ans %= 998244353;
  }
  dizdp[nokta] = ans;
  return ans;
}
long long fak(long long a) {
  assert(a < 20000);
  if (a < 2) return 1;
  if (fakdp[a] != -1) return fakdp[a];
  long long ans = fak(a - 1) * a;
  ans %= 998244353;
  fakdp[a] = ans;
  return ans;
}
long long comb(long long a, long long b) {
  if (a < b) return 0;
  long long payda = fak(b) * fak(a - b);
  payda %= 998244353;
  long long ans = fak(a) * us(payda, 998244353 - 2);
  ans %= 998244353;
  return ans;
}
long long cp(long long a, long long b) {
  long long ans = a * b;
  ans %= 998244353;
  return ans;
}
int main() {
  clock_t start = clock();
  cin >> n >> k >> l;
  long long seglen = l * us(2 * n + 1, 998244353 - 2);
  seglen %= 998244353;
  for (int i = 0; i < 20000; i++) {
    fakdp[i] = -1;
    dizdp[i] = -1;
  }
  long long save = 0;
  long long ans = 0;
  for (int i = k; i < n; i++) {
    long long sol = i;
    long long sag = 2 * n - i;
    for (int j = k; j <= sol && j <= sag; j++) {
      if ((sol - j) % 2 != 0) continue;
      long long ekle = comb(sol, j) * comb(sag, j);
      ekle %= 998244353;
      ekle *= fak(j);
      ekle %= 998244353;
      ekle *= diz(sol - j);
      ekle %= 998244353;
      ekle *= diz(sag - j);
      ekle %= 998244353;
      ans += ekle;
      ans %= 998244353;
    }
  }
  save = ans;
  {
    int i = n;
    long long sol = i;
    long long sag = 2 * n - i;
    for (int j = k; j <= sol && j <= sag; j++) {
      if ((sol - j) % 2 != 0) continue;
      long long ekle = comb(sol, j) * comb(sag, j);
      ekle %= 998244353;
      ekle *= fak(j);
      ekle %= 998244353;
      ekle *= diz(sol - j);
      ekle %= 998244353;
      ekle *= diz(sag - j);
      ekle %= 998244353;
      ans += ekle;
      ans %= 998244353;
    }
  }
  ans += save;
  ans %= 998244353;
  ans *= us(diz(2 * n), 998244353 - 2);
  ans %= 998244353;
  ans *= seglen;
  ans %= 998244353;
  cout << ans << endl;
  cerr << (clock() - start) / (double)CLOCKS_PER_SEC << "ms" << endl;
}