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 = 2005, P = 998244353;
int n, k, l, ans, fac[N], ifac[N], inv[N << 1], f[N << 1];
inline int Pow(long long x, int y = P - 2) {
  int ans = 1;
  for (; y; y >>= 1, x = x * x % P)
    if (y & 1) ans = ans * x % P;
  return ans;
}
inline int calc() {
  int now = 0;
  for (int i = 0; i <= n * 2; ++i)
    now = (now + (long long)f[i] * inv[i + 1]) % P;
  return now;
}
int main() {
  scanf("%d%d%d", &n, &k, &l);
  inv[1] = 1;
  for (int i = 2; i <= n * 2 + 1; ++i)
    inv[i] = (long long)(P - P / i) * inv[P % i] % P;
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = (long long)fac[i - 1] * i % P;
  ifac[n] = Pow(fac[n]);
  for (int i = n; i; --i) ifac[i - 1] = (long long)ifac[i] * i % P;
  f[0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 2 * n; j > 1; --j)
      f[j] = ((P - 2ll) * f[j - 2] + 2ll * f[j - 1]) % P;
    f[1] = (2ll * f[0]) % P;
    f[0] = 0;
  }
  for (int i = n; i >= k; --i) {
    ans =
        (ans + (long long)fac[n] * ifac[i] % P * ifac[n - i] % P * calc()) % P;
    memmove(f, f + 1, (n * 2 + 1) << 2);
    for (int j = 0; j <= n * 2; ++j) f[j] = f[j] * (P + 1ll) / 2 % P;
    for (int j = 1; j <= n * 2; ++j) (f[j] += f[j - 1]) %= P;
    f[n * 2] = f[n * 2 - 1] = 0;
    for (int j = 2 * n; j > 1; --j)
      f[j] = (2ll * f[j - 2] + (P - 2ll) * f[j - 1] + f[j]) % P;
    f[1] = ((P - 2ll) * f[0] + f[1]) % P;
  }
  return printf("%lld", (long long)ans * l % P), 0;
}