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;
long long dp[4005][2005][2];
const int fish = 998244353;
long long Pow(long long x, int y) {
  long long ans = 1, b = x;
  while (y != 0) {
    if (y % 2) ans = (1ll * ans * b) % fish;
    y /= 2;
    b = (1ll * b * b) % fish;
  }
  return ans;
}
int main() {
  scanf("%d%d%d", &N, &K, &L);
  dp[0][0][0] = 1;
  for (int i = 1; i <= 2 * N + 1; i++)
    for (int j = 0; j <= N; j++)
      for (int k = 0; k <= 1; k++) {
        if (!k && j >= K) dp[i][j][1] = (dp[i][j][1] + dp[i - 1][j][0]) % fish;
        if (i + j - 1 < 2 * N + k && j - 1 >= 0) {
          dp[i][j][k] = (dp[i][j][k] + dp[i - 1][j - 1][k]) % fish;
        }
        dp[i][j][k] =
            (dp[i][j][k] + 1ll * dp[i - 1][j + 1][k] * (j + 1) % fish) % fish;
      }
  long long fac = 1;
  for (int i = N + 1; i <= 2 * N + 1; i++) fac = (1ll * fac * i) % fish;
  long long Ny = Pow(fac, fish - 2);
  long long ans = 1ll * dp[2 * N + 1][0][1] * 1ll;
  ans = 1ll * ans * Ny % fish;
  ans = 1ll * ans * L % fish;
  ans = ans * Pow(2, N) % fish;
  cout << ans;
  return 0;
}