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 MX = 2005, mod = 998244353;
int n, k, l;
long long mem[2 * MX][MX][2];
long long pot(long long b, int p) {
  long long res = 1;
  while (p) {
    if (p & 1) (res *= b) %= mod;
    (b *= b) %= mod;
    p /= 2;
  }
  return res;
}
long long dp(int i, int j, int f) {
  if (j > n) return 0;
  if (i == 2 * n + 1) return f && !j;
  long long &res = mem[i][j][f];
  if (res != -1) return res;
  res = dp(i + 1, j + 1, f);
  if (j) res += j * dp(i + 1, j - 1, f);
  if (j >= k && !f) res += dp(i + 1, j, 1);
  return res %= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> l;
  memset(mem, -1, sizeof(mem));
  long long p = dp(0, 0, 0);
  k = 0;
  memset(mem, -1, sizeof(mem));
  long long q = dp(0, 0, 0);
  cout << l * p % mod * pot(q, mod - 2) % mod << '\n';
  return 0;
}