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 mul(long long a, long long b) {
  if (!b) return 1;
  long long k = mul(a, b / 2);
  if (b & 1) return k * k % 998244353 * a % 998244353;
  return k * k % 998244353;
}
int n, k, l;
long long fac[5010], invfac[5010], mul2[5010], dp[5010];
void init(int lim) {
  fac[0] = 1, invfac[0] = 1, mul2[0] = 1;
  for (int i = 1; i <= lim; i++) {
    fac[i] = fac[i - 1] * i % 998244353;
    invfac[i] = mul(fac[i], 998244353 - 2);
    mul2[i] = mul2[i - 1] * 2 % 998244353;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k >> l;
  init(5000);
  for (int i = n; i >= k; i--) {
    dp[i] = fac[i] * fac[i] % 998244353 * mul2[i] % 998244353 *
            invfac[2 * i + 1] % 998244353;
    dp[i] *=
        fac[n] % 998244353 * invfac[i] % 998244353 * invfac[n - i] % 998244353;
    for (int j = i + 1; j <= n; j++)
      (dp[i] -= dp[j] * fac[j] % 998244353 * invfac[i] % 998244353 *
                invfac[j - i] % 998244353) %= 998244353;
    dp[i] = (dp[i] + 998244353) % 998244353;
  }
  long long ans = 0;
  for (int i = k; i <= n; i++) (ans += dp[i]) %= 998244353;
  cout << (ans * l) % 998244353 << endl;
  return 0;
}