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 N = 4005;
int f[N][N][2], cheng[N];
int n, m;
const int mod = 998244353;
long long poww(long long x, long long nn) {
  long long res = 1;
  while (nn) {
    if (nn & 1) res = res * x % mod;
    x = x * x % mod;
    nn >>= 1;
  }
  return res;
}
long long l;
int main() {
  cin >> n >> m >> l;
  f[0][0][0] = 1;
  for (int i = 0; i <= 2 * n; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k < 2; k++) {
        (f[i + 1][j + 1][k] += f[i][j][k]) %= mod;
        if (j) (f[i + 1][j - 1][k] += 1ll * f[i][j][k] * j % mod) %= mod;
        if (!k && j >= m) (f[i + 1][j][1] += f[i][j][k]) %= mod;
      }
    }
  }
  cheng[0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++) cheng[i] = 1ll * cheng[i - 1] * i % mod;
  printf("%lld\n", l * f[2 * n + 1][0][1] % mod * poww(2, n) % mod * cheng[n] %
                       mod * poww(cheng[2 * n + 1], mod - 2) % mod);
}