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 int read() {
  int s = 0, t = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * t;
}
const int N = 4005;
const int P = 998244353;
int n, K, L, IL, Ans;
int fac[N], inv[N], coe[N], f[N], g[N];
int Mod(int x) { return x >= P ? x - P : x; }
int Ksm(int x, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * x % P;
    x = 1ll * x * x % P, k >>= 1;
  }
  return res;
}
void Pre(int n) {
  coe[0] = 1;
  for (int i = 1; i <= n; i++) coe[i] = 1ll * coe[i - 1] * L % P;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % P;
  inv[n] = Ksm(fac[n], P - 2);
  for (int i = n; i >= 1; i--) inv[i - 1] = 1ll * inv[i] * i % P;
}
int C(int n, int m) { return 1ll * fac[n] * inv[m] % P * inv[n - m] % P; }
int main() {
  n = read(), K = read(), L = read(), Pre(N - 1);
  f[0] = 1, IL = Ksm(L, P - 2);
  for (int i = 1, m = 2; i <= n; i++, m += 2) {
    int a = 2ll * IL % P, b = Mod(P - 2ll * IL * IL % P);
    for (int j = m; j >= 1; j--) {
      if (j > 1)
        f[j] = Mod(1ll * a * f[j - 1] % P + 1ll * b * f[j - 2] % P);
      else
        f[j] = 1ll * a * f[j - 1] % P;
    }
    f[0] = 0;
    int res = 0;
    for (int j = 0; j <= m; j++) {
      int t = 1ll * inv[j + 1] * fac[j] % P;
      res = Mod(res + 1ll * t * coe[j + 1] % P * f[j] % P);
    }
    g[i] = 1ll * res * C(n, i) % P;
  }
  memset(f, 0, sizeof(f));
  for (int i = 0; i <= n; i++)
    for (int j = i; j <= n; j++)
      f[i] =
          Mod(f[i] + 1ll * C(j, i) * ((j - i) & 1 ? P - 1 : 1) % P * g[j] % P);
  for (int i = K; i <= n; i++) Ans = Mod(Ans + f[i]);
  cout << Ans;
  return 0;
}