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>
const long long mod = 998244353;
long long n, m, l, dp[4010][4010][2], fac[4010];
long long quickpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y % 2 == 0) return quickpow(x * x % mod, y / 2);
  return quickpow(x * x % mod, y / 2) * x % mod;
}
signed main() {
  scanf("%lld%lld%lld", &n, &m, &l);
  dp[0][0][0] = 1;
  for (long long i = 0; i <= 2 * n; i++)
    for (long long j = 0; j <= i; j++)
      for (long long o = 0; o <= 1; o++) {
        (dp[i + 1][j + 1][o] += dp[i][j][o]) %= mod;
        if (j >= m && !o) (dp[i + 1][j][1] += dp[i][j][0]);
        if (j) (dp[i + 1][j - 1][o] += dp[i][j][o] * j % mod) %= mod;
      }
  long long ans = dp[2 * n + 1][0][1];
  fac[0] = 1;
  for (long long i = 1; i <= 2 * n + 1; i++) fac[i] = fac[i - 1] * i % mod;
  (ans *= quickpow(2, n) * fac[n] % mod * quickpow(fac[n * 2 + 1], mod - 2) %
          mod * l % mod) %= mod;
  printf("%lld\n", ans);
}