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 read() {
  long long u = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    u = u * 10 + c - '0';
    c = getchar();
  }
  return u * f;
}
inline long long pw(long long x, long long y) {
  long long res = 1;
  for (; y; y >>= 1, x = x * x % 998244353) {
    if (y & 1) res = res * x % 998244353;
  }
  return res;
}
long long l, n, k, f[2005 << 1][2005], g[2005 << 1][2005];
signed main() {
  n = read();
  k = read();
  l = read();
  f[0][0] = 1;
  for (long long i = 1; i <= 2 * n; i++) {
    for (long long j = 0; j <= min(i, n); j++) {
      if (!j)
        f[i][j] = f[i - 1][j + 1], g[i][j] = g[i - 1][j + 1];
      else
        f[i][j] =
            (f[i - 1][j - 1] + 1ll * (j + 1) * f[i - 1][j + 1]) % 998244353,
        g[i][j] =
            (g[i - 1][j - 1] + 1ll * (j + 1) * g[i - 1][j + 1]) % 998244353;
      if (j >= k) g[i][j] = (f[i][j] + g[i][j]) % 998244353;
    }
  }
  cout << (l * pw(f[2 * n][0] * (2 * n + 1) % 998244353, 998244353 - 2) %
           998244353 * g[2 * n][0] % 998244353)
       << endl;
  return 0;
}