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 int read() {
  int w = 1, s = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return w * s;
}
const int mod = 998244353;
int n, L, k, f[4040][2020][2], fac[4040];
inline long long ksm(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
inline int pls(int a, int b) {
  a += b;
  return a >= mod ? a - mod : a;
}
int main() {
  n = read(), k = read(), L = read();
  f[0][0][0] = 1;
  int N = 2 * n + 1;
  for (register int i = 0; i < N; ++i) {
    for (register int j = 0; j <= n; ++j) {
      for (register int t = 0; t < 2; ++t) {
        if (!f[i][j][t]) continue;
        if (!t && j >= k) f[i + 1][j][1] = pls(f[i + 1][j][1], f[i][j][t]);
        if (i + j + 2 <= 2 * n + t)
          f[i + 1][j + 1][t] = pls(f[i + 1][j + 1][t], f[i][j][t]);
        if (j)
          f[i + 1][j - 1][t] =
              pls(f[i + 1][j - 1][t], 1ll * f[i][j][t] * j % mod);
      }
    }
  }
  fac[0] = 1;
  for (register int i = 1; i <= N; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  long long tmp = 1ll * fac[n] * ksm(2, n) % mod * f[N][0][1] % mod *
                  ksm(fac[N], mod - 2) % mod;
  cout << tmp * L % mod;
  return 0;
}