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 int N = 4005;
const int mod = 998244353;
template <typename T>
void read(T &x) {
  int f = 0;
  register char c = getchar();
  while (c < '0' || c > '9') f |= (c == '-'), c = getchar();
  for (x = 0; c >= '0' && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + (c ^ '0');
  if (f) x = -x;
}
int Qpow(int x, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) ans = 1LL * ans * x % mod;
    x = 1LL * x * x % mod;
    p >>= 1;
  }
  return ans;
}
int Inv(int x) { return Qpow(x, mod - 2); }
int n, k, l;
int fac[N], f[N][N];
void upd(int &x, register int y) {
  static int z;
  x = ((z = x + y) >= mod) ? (z - mod) : z;
}
int main() {
  read(n);
  read(k);
  read(l);
  f[0][0] = 1;
  for (int i = 1; i <= n * 2; ++i) {
    for (int j = 0; j <= n && j <= i; ++j) {
      upd(f[i][j], 1LL * f[i - 1][j + 1] * (j + 1) % mod);
      if (j) upd(f[i][j], f[i - 1][j - 1]);
    }
  }
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = 1LL * fac[i - 1] * i % mod;
  }
  int ans = 0;
  for (int i = 1; i < 2 * n; ++i) {
    for (int j = k; j <= n && j <= i; ++j) {
      upd(ans, 1LL * f[i][j] * f[2 * n - i][j] % mod * fac[j] % mod);
    }
  }
  ans = 1LL * ans * l % mod * Inv(2 * n + 1) % mod;
  ans = 1LL * ans * Inv(f[2 * n][0]) % mod;
  cout << ans << endl;
  return 0;
}