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 int K = 998244353;
long long int mu(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int q = mu(a, n / 2);
  if (n % 2 == 0) return q * q % K;
  return q * q % K * a % K;
}
long long int d[4005][2005];
long long int gt[4005], rev2[4005];
int n, k;
long long int calc2(int u1, int u0) { return gt[u1 + 2 * u0] * rev2[u0] % K; }
int cnt = 0;
long long int calc(int i, int u1) {
  if (d[i][u1] != -1) return d[i][u1];
  if (i == n * 2) return 0;
  cnt++;
  int u2 = (i - u1) / 2;
  int u0 = (n - u1 - u2);
  long long int res = 0;
  if (u0 >= 1) {
    long long int u = calc(i + 1, u1 + 1);
    res = (res + u0 * 1ll * u % K) % K;
  }
  if (u1 >= 1) {
    long long int u = calc(i + 1, u1 - 1);
    res = (res + u1 * 1ll * u % K) % K;
  }
  if (u1 >= k) res = (res + calc2(u1, u0)) % K;
  return d[i][u1] = res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  long long int l;
  cin >> n >> k >> l;
  memset(d, -1, sizeof(d));
  gt[0] = 1;
  rev2[0] = 1;
  long long int pw2 = 1;
  for (int i = 1; i <= 4000; i++) {
    gt[i] = gt[i - 1] * 1ll * i % K;
    pw2 = pw2 * 2 % K;
    rev2[i] = mu(pw2, K - 2) % K;
  }
  long long int res = calc(0, 0) * mu(calc2(0, n), K - 2) % K;
  res = (res * l % K * mu(2 * n + 1, K - 2) % K);
  cout << res << endl;
  return 0;
}