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;
long long pw(long long a, long long b) {
  long long c = a, d = 1;
  while (b) {
    if (b & 1) {
      d *= c;
      d %= 998244353LL;
    }
    c *= c;
    c %= 998244353LL;
    b /= 2;
  }
  return d;
}
inline long long dv(long long a, long long b) {
  a %= 998244353LL;
  b %= 998244353LL;
  return (a * pw(b, 998244353LL - 2)) % 998244353LL;
}
long long fct[11234] = {1};
long long pw2[11234] = {1};
long long bet[11234];
inline long long cmb(long long a, long long b) {
  return dv(fct[a], fct[b] * fct[a - b]);
}
int main() {
  for (int i = (1); i < (11234); i++) fct[i] = (fct[i - 1] * i) % 998244353LL;
  for (int i = (1); i < (11234); i++) pw2[i] = (pw2[i - 1] * 2) % 998244353LL;
  for (int i = (0); i < ((11234)); i++)
    bet[i] = dv(fct[i] * fct[i], fct[2 * i + 1]);
  ;
  long long n, k, l;
  scanf("%lld%lld%lld", &n, &k, &l);
  long long ret = 0;
  for (int j = (k); j < (n + 1); j++) {
    long long a = cmb(n, j);
    for (int l = (0); l < ((n - j + 1)); l++) {
      long long b = cmb(n - j, l);
      if (l % 2) b = (-b + 998244353LL * 3) % 998244353LL;
      b *= pw2[j + l];
      b %= 998244353LL;
      b *= bet[j + l];
      b %= 998244353LL;
      ret += a * b;
      ret %= 998244353LL;
    }
  }
  ret *= l;
  ret %= 998244353LL;
  printf("%lld\n", ret);
}