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 int mod = 998244353;
int n, m, i, j, k, l, ans, f[4010][2010][2], inv[4010];
int main() {
  scanf("%d%d%d", &n, &m, &l);
  f[0][0][0] = 1;
  for (int i = 0; i <= n * 2; i++)
    for (int j = 0; j <= n; j++)
      for (int k = 0; k < 2; k++)
        if (f[i][j][k]) {
          f[i + 1][j + 1][k] = (f[i + 1][j + 1][k] + f[i][j][k]) % mod;
          if (j)
            f[i + 1][j - 1][k] =
                (f[i + 1][j - 1][k] + (long long)f[i][j][k] * j) % mod;
          if (!k && j >= m)
            f[i + 1][j][1] = (f[i + 1][j][1] + f[i][j][0]) % mod;
        }
  ans = f[2 * n + 1][0][1], inv[1] = 1;
  for (int i = 2; i <= 2 * n + 1; i++) {
    inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod;
    ans = (long long)ans * inv[i] % mod;
  }
  for (int i = 1; i <= n; i++) ans = (long long)ans * 2 % mod * i % mod;
  printf("%d\n", (long long)ans * l % mod);
  return 0;
}