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 int n, l, k;
long long int fact[4002];
long long int dp[4002][2002][2];
long long int f(long long int i, long long int j, long long int x) {
  if (i == n + n + 1) return x;
  if (dp[i][j][x] != -1) return dp[i][j][x];
  long long int total = 0;
  if (x == 0 && j >= k) total += f(i + 1, j, 1);
  total %= 998244353;
  if ((n + n + 1) - i > j + (1 - x)) total += f(i + 1, j + 1, x);
  total %= 998244353;
  if (j > 0) total += 1ll * ((j * f(i + 1, j - 1, x)) % 998244353);
  total %= 998244353;
  return dp[i][j][x] = total;
}
long long int exp(long long int n, long long int e) {
  if (e == 0) return 1;
  if ((e & 1) == 0) return exp((n * n) % 998244353, e >> 1);
  return (n * exp(n, e - 1)) % 998244353;
}
int main() {
  fact[0] = 1;
  cin >> n >> k >> l;
  for (int i = 1; i <= n + n + 1; i++) fact[i] = (i * fact[i - 1]) % 998244353;
  for (int i = 0; i <= n + n + 1; i++)
    for (int j = 0; j <= n; j++) dp[i][j][0] = dp[i][j][1] = -1ll;
  long long int ans = f(0, 0, 0);
  ans = ans * exp(fact[n + n + 1], 998244353 - 2);
  ans %= 998244353;
  ans = ans * fact[n];
  ans %= 998244353;
  ans = ans * exp(2, n);
  ans %= 998244353;
  ans = ans * l;
  ans = ans % 998244353;
  cout << ans << "\n";
  return 0;
}