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 long long mod = 998244353;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
int m, n, x, t, y, k, l, r, z, all, ans;
long long b[4005];
long long f[4005][2005][2];
char c;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n, long long p = (long long)998244353) {
  long long r = 1 % p;
  for (a %= p; n; a = a * a % p, n >>= 1)
    if (n & 1) r = r * a % p;
  return r;
}
long long inv(long long x, long long p = (long long)998244353) {
  return x <= 1 ? 1 : inv(p % x, p) * (p - p / x) % p;
}
void init() {}
void solve() {
  scanf("%d%d%d", &n, &k, &l);
  f[1][1][0] = 1;
  b[1] = 1;
  for (int i = 2; i <= 2 * n + 1; i++) {
    for (int j = 0; j <= i && j <= n; j++) {
      f[i][j][0] =
          ((j > 0 ? f[i - 1][j - 1][0] : 0) + f[i - 1][j + 1][0] * (j + 1)) %
          mod;
      f[i][j][1] =
          ((j > 0 ? f[i - 1][j - 1][1] : 0) + f[i - 1][j + 1][1] * (j + 1) +
           (j >= k ? f[i - 1][j][0] : 0)) %
          mod;
    }
    b[i] = b[i - 1] * i % mod;
  }
  long long ans = f[2 * n + 1][0][1] * qpow(2, n) % mod * b[n] % mod * l % mod *
                  inv(b[2 * n + 1]) % mod;
  printf("%lld\n", ans);
}
int main() {
  int T;
  T = 1;
  for (int i = 1; i <= T; i++) {
    init();
    solve();
  }
  return 0;
}