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 INF = 1e9 + 10;
const int SZ = 1e6 + 10;
const int mod = 998244353;
const double PI = acos(-1);
const double eps = 1e-7;
long long read() {
  long long n = 0;
  char a = getchar();
  bool flag = 0;
  while (a > '9' || a < '0') {
    if (a == '-') flag = 1;
    a = getchar();
  }
  while (a <= '9' && a >= '0') {
    n = n * 10 + a - '0', a = getchar();
  }
  if (flag) n = -n;
  return n;
}
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = a * ans % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int f[4010][4010][2];
int getfac(int x) {
  int ans = 1;
  for (int i = 2; i <= x; i++) ans = 1ll * ans * i % mod;
  return ans;
}
int main() {
  int n = read(), m = read(), l = read();
  f[0][0][0] = 1;
  for (int i = 0; i < 2 * n + 1; i++) {
    for (int j = 0; j <= i; j++) {
      if (j >= m) (f[i + 1][j][1] += f[i][j][0]) %= mod;
      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;
      }
    }
  }
  int ans = f[2 * n + 1][0][1];
  ans = 1ll * ans * getfac(n) % mod * ksm(2, n) % mod *
        ksm(getfac(2 * n + 1), mod - 2) % mod;
  ans = 1ll * ans * l % mod;
  cout << ans << endl;
}