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>
#pragma GCC optimize("O2")
using namespace std;
long long powi(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1, a = a * a % 998244353LL) {
    if (b & 1) ret = ret * a % 998244353LL;
  }
  return ret;
}
long long inv(long long a) { return powi(a, 998244353LL - 2); }
long long f[5000] = {1}, invf[5000] = {1}, ff[5000] = {1};
long long C(long long n, long long m) {
  return f[n] * invf[m] % 998244353LL * invf[n - m] % 998244353LL;
}
long long C(long long n) { return ff[n]; }
int main() {
  cin.tie(0), ios::sync_with_stdio(0);
  for (int i = 1; i < 5000; i++) {
    f[i] = f[i - 1] * i % 998244353LL;
    invf[i] = inv(f[i]);
    ff[i] = ff[i - 1] * (i * 2 - 1) % 998244353LL;
  }
  long long n, k, l;
  cin >> n >> k >> l;
  long long q = (n + n + 1) * C(n) % 998244353LL, p = 0;
  for (int i = 0; i <= n + n; i++) {
    int l = i, r = n + n - i;
    for (int j = k; j <= n; j++) {
      if (l < j || r < j) continue;
      if (((l - j) & 1) || ((r - j) & 1)) continue;
      p = (p + C(l, j) * C(r, j) % 998244353LL * f[j] % 998244353LL *
                   C((l - j) / 2) % 998244353LL * C((r - j) / 2)) %
          998244353LL;
    }
  }
  cout << l * p % 998244353LL * inv(q) % 998244353LL << '\n';
  return 0;
}