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;
inline long long getint() {
  long long num = 0, flag = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag = -1;
  while (c >= '0' && c <= '9') num = num * 10 + c - 48, c = getchar();
  return num * flag;
}
int n, k, l;
int f[5005][5005][2], fac[5005];
inline int upd(int x) { return x < 998244353 ? x : x - 998244353; }
inline int ksm(int num, int k) {
  int ret = 1;
  for (; k; k >>= 1, num = 1ll * num * num % 998244353)
    if (k & 1) ret = 1ll * ret * num % 998244353;
  return ret;
}
int main() {
  n = getint(), k = getint(), l = getint();
  f[0][0][0] = 1;
  for (int i = 0; i <= n * 2; i++)
    for (int j = 0; j <= i; j++)
      for (int o = 0; o <= 1; o++) {
        f[i + 1][j + 1][o] = upd(f[i + 1][j + 1][o] + f[i][j][o]);
        if (!o && j >= k) f[i + 1][j][1] = upd(f[i + 1][j][1] + f[i][j][0]);
        if (j)
          f[i + 1][j - 1][o] =
              upd(f[i + 1][j - 1][o] + 1ll * f[i][j][o] * j % 998244353);
      }
  fac[0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++)
    fac[i] = 1ll * fac[i - 1] * i % 998244353;
  printf("%lld\n", 1ll * f[2 * n + 1][0][1] * fac[n] % 998244353 * ksm(2, n) %
                       998244353 * ksm(fac[2 * n + 1], 998244353 - 2) %
                       998244353 * l % 998244353);
}