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 md = 998244353;
const int xn = 2e5 + 10;
const int xm = -20 + 10;
const int SQ = 320;
const int sq = 320 + 10;
const int inf = 1e9 + 10;
const long long INF = 1e10 + 10;
long long power(long long a, long long b) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2) % md
                      : power(a * a % md, b / 2) % md));
}
long long zarb(long long a, long long b) { return (a * b + 10 * md) % md; }
long long jaam(long long a, long long b) { return (a + b + 10 * md) % md; }
string abc = "abcdefghijklmnopqrstuvwxyz";
long long n, k, l, dp[xn], fact[xn], ifact[xn], ans, inv, res;
long long C(long long r, long long n) {
  if (r < 0 || r > n || n <= 0) return 0;
  return fact[n] * (ifact[r] * ifact[n - r] % md) % md;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k >> l;
  dp[0] = 1;
  for (long long i = 2; i < xn; i += 2) dp[i] = (dp[i - 2] * (i - 1) % md);
  fact[0] = ifact[0] = 1;
  for (long long i = 1; i < xn; ++i) {
    fact[i] = fact[i - 1] * i % md;
    ifact[i] = power(fact[i], md - 2);
  }
  for (long long i = 1; i < n + n; ++i) {
    for (long long j = k; j <= min(i, n + n - i); ++j) {
      res = C(j, i) * C(j, n + n - i) % md;
      res = res * (dp[i - j] * dp[n + n - i - j] % md) % md;
      res = res * fact[j] % md;
      ans = (ans + res) % md;
    }
  }
  ans = ans * power(dp[n + n], md - 2) % md;
  ans = ans * (l * power(n + n + 1, md - 2) % md) % md;
  cout << ans << '\n';
  return 0;
}