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;
int n, k, l, DP[5003][5003][2];
int _pow(int a, int b) {
  if (b == 1) return a;
  if (b % 2 == 0) return _pow(1LL * a * a % 998244353, b / 2);
  return 1LL * _pow(1LL * a * a % 998244353, b / 2) * a % 998244353;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k >> l;
  DP[0][0][0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++) {
    for (int j = 0; j <= min(i, n); j++) {
      if (j > 0) {
        if (i + j <= 2 * n) DP[i][j][0] = DP[i - 1][j - 1][0];
        if (i + j <= 2 * n + 1) DP[i][j][1] = DP[i - 1][j - 1][1];
      }
      DP[i][j][0] =
          (DP[i][j][0] + 1LL * DP[i - 1][j + 1][0] * (j + 1)) % 998244353;
      DP[i][j][1] =
          (DP[i][j][1] + 1LL * DP[i - 1][j + 1][1] * (j + 1)) % 998244353;
      if (j >= k) DP[i][j][1] = (DP[i][j][1] + DP[i - 1][j][0]) % 998244353;
    }
  }
  int ans = DP[2 * n + 1][0][1], x = 1;
  for (int i = 1; i <= n; i++) x = 1LL * x * i % 998244353;
  ans = 1LL * ans * x % 998244353;
  x = 1;
  for (int i = 1; i <= 2 * n + 1; i++) x = 1LL * x * i % 998244353;
  ans = 1LL * ans * _pow(x, 998244353 - 2) % 998244353;
  ans = 1LL * ans * _pow(2, n) % 998244353;
  cout << 1LL * ans * l % 998244353 << '\n';
  return 0;
}