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 Getint() {
  char ch = getchar();
  long long x = 0, fh = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    (x *= 10) += ch ^ 48;
    ch = getchar();
  }
  return x * fh;
}
const long long Mod = 998244353;
const int N = 4025;
int K, n, l;
long long f[N][N], fc[N], fiv[N], inv[N];
inline long long Ksm(long long x, long long y) {
  long long tmp = 1;
  while (y) {
    if (y & 1) tmp = tmp * x % Mod;
    y >>= 1;
    if (y) x = x * x % Mod;
  }
  return tmp;
}
int main() {
  n = Getint();
  K = Getint();
  l = Getint();
  fc[0] = fc[1] = fiv[0] = fiv[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= N - 1; i++) {
    fc[i] = fc[i - 1] * i % Mod;
    inv[i] = (Mod - Mod / i) * inv[Mod % i] % Mod;
    fiv[i] = fiv[i - 1] * inv[i] % Mod;
  }
  f[0][0] = 1;
  for (int i = 0; i <= 2 * n - 1; i++)
    for (int j = 0; j <= i; j++) {
      long long val = f[i][j];
      if (!val) continue;
      if (j > 0) (f[i + 1][j - 1] += val * j) %= Mod;
      (f[i + 1][j + 1] += val) %= Mod;
    }
  long long Ans = 0;
  for (int i = 2; i <= 2 * n; i++)
    for (int k = K; k <= 2 * n; k++)
      (Ans += i * f[2 * n - i + 1][k] % Mod * f[i - 1][k] % Mod * fc[k]) %= Mod;
  for (int i = 1; i <= 2 * n - 1; i++)
    for (int k = K; k <= 2 * n; k++)
      (Ans -= i * f[i][k] % Mod * f[2 * n - i][k] % Mod * fc[k]) %= Mod;
  Ans = Ans * inv[2 * n + 1] % Mod * Ksm(f[2 * n][0], Mod - 2) % Mod * l % Mod;
  cout << (Ans % Mod + Mod) % Mod << '\n';
  return 0;
}