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>
const int mod = 998244353;
const int N = 262144;
using LL = long long;
int factor[N], ifactor[N], n, k, l;
int lim, s, rev[N], wn[N], w[N], ans;
void reduce(int &x) { x += x >> 31 & mod; }
int pow(int x, int y, int ans = 1) {
  for (; y; y >>= 1, x = (LL)x * x % mod)
    if (y & 1) ans = (LL)ans * x % mod;
  return ans;
}
void fftinit(int len) {
  wn[0] = lim = 1, s = -1;
  while (lim < len) lim <<= 1, ++s;
  for (int i = 0; i < lim; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
  const int g = pow(3, (mod - 1) / lim);
  for (int i = 1; i < lim; ++i) wn[i] = (LL)wn[i - 1] * g % mod;
}
void fft(int *A, int typ) {
  for (int i = 0; i < lim; ++i)
    if (i < rev[i]) std::swap(A[i], A[rev[i]]);
  for (int i = 1; i < lim; i <<= 1) {
    for (int j = 0, t = lim / i / 2; j < i; ++j) w[j] = wn[j * t];
    for (int j = 0; j < lim; j += i << 1)
      for (int k = 0; k < i; ++k) {
        const int x = A[k + j], y = (LL)A[k + j + i] * w[k] % mod;
        reduce(A[k + j] += y - mod), reduce(A[k + j + i] = x - y);
      }
  }
  if (!typ) {
    const int il = pow(lim, mod - 2);
    std::reverse(A + 1, A + lim);
    for (int i = 0; i < lim; ++i) A[i] = (LL)A[i] * il % mod;
  }
}
int A[N], B[N], C[N];
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0);
  std::cin >> n >> k >> l;
  factor[0] = 1;
  for (int i = 1; i <= n * 2 + 1; ++i) factor[i] = (LL)factor[i - 1] * i % mod;
  ifactor[n] = pow(factor[n], mod - 2);
  for (int i = n; i; --i) ifactor[i - 1] = (LL)ifactor[i] * i % mod;
  fftinit(n + n + 1);
  for (int i = k; i <= n; ++i) A[i] = ifactor[i];
  for (int i = 0; i <= n; ++i) B[i] = i & 1 ? mod - ifactor[i] : ifactor[i];
  fft(A, 1), fft(B, 1);
  for (int i = 0; i < lim; ++i) C[i] = (LL)A[i] * B[i] % mod;
  fft(C, 0);
  for (int i = k; i <= n; ++i)
    reduce(ans += (LL)pow(2, i) * pow(factor[i], 2) % mod *
                      pow(factor[2 * i + 1], mod - 2) % mod * C[i] % mod *
                      ifactor[n - i] % mod -
                  mod);
  std::cout << (LL)ans * l % mod * factor[n] % mod << '\n';
  return 0;
}