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 read() {
  long long x = 0, f = 1;
  char p = getchar();
  while (!isdigit(p)) {
    if (p == '-') f = -1;
    p = getchar();
  }
  while (isdigit(p)) x = (x << 3) + (x << 1) + (p ^ 48), p = getchar();
  return x * f;
}
const long long maxn = 8e3 + 5, mod = 998244353, G = 3, Gi = 332748118;
long long f[maxn], g[maxn], h[maxn], n, k, l, inv[maxn], cheng[maxn],
    len = 1, rev[maxn], ans;
inline long long power(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
inline void ntt(long long *f, long long n, long long op) {
  for (long long i = 0; i < n; i++)
    if (i < rev[i]) swap(f[i], f[rev[i]]);
  for (long long i = 1; i < n; i <<= 1) {
    long long wn;
    if (op == 1)
      wn = power(G, (mod - 1) / (i << 1));
    else
      wn = power(Gi, (mod - 1) / (i << 1));
    for (long long j = 0; j < n; j += (i << 1)) {
      long long w = 1;
      for (long long k = 0; k < i; k++, w = w * wn % mod) {
        long long nx = f[j + k], ny = f[i + j + k] * w % mod;
        f[j + k] = (nx + ny) % mod;
        f[i + j + k] = (nx - ny + mod) % mod;
      }
    }
  }
  if (op == 1) return;
  long long iv = power(n, mod - 2);
  for (long long i = 0; i < n; i++) f[i] = f[i] * iv % mod;
}
signed main() {
  n = read();
  k = read();
  l = read();
  cheng[0] = 1;
  for (long long i = 1; i <= 2 * n + 1; i++) cheng[i] = cheng[i - 1] * i % mod;
  inv[0] = inv[1] = 1;
  inv[2 * n + 1] = power(cheng[2 * n + 1], mod - 2);
  for (long long i = 2 * n; i >= 2; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  for (long long i = k; i <= n; i++) f[i] = inv[i];
  g[0] = 1;
  for (long long i = 1; i <= n; i++) g[i] = mod - g[i - 1];
  for (long long i = 1; i <= n; i++) g[i] = g[i] * inv[i] % mod;
  h[0] = 1;
  for (long long i = 1; i <= n; i++) h[i] = h[i - 1] * 2 % mod;
  for (long long i = 0; i <= n; i++)
    h[i] = h[i] * cheng[i] % mod * cheng[i] % mod * inv[2 * i + 1] % mod *
           inv[n - i] % mod;
  while (len <= n + n) len <<= 1;
  for (long long i = 1; i < len; i++)
    rev[i] = ((rev[i >> 1] >> 1) | ((i & 1) ? (len >> 1) : 0));
  ntt(f, len, 1);
  ntt(g, len, 1);
  for (long long i = 0; i < len; i++) f[i] = f[i] * g[i] % mod;
  ntt(f, len, -1);
  for (long long i = 0; i <= n; i++) h[i] = h[i] * f[i] % mod;
  for (long long i = k; i <= n; i++) ans = (ans + h[i]) % mod;
  ans = ans * cheng[n] % mod * l % mod;
  cout << ans << endl;
  return 0;
}