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 ll = long long;
const int C = 5001, M = 998244353;
inline ll pow(ll a, ll b, ll M) {
  if (b == 0) return 1;
  ll c = 1;
  while (b > 0) {
    if (b % 2 == 1) c = (c * a) % M;
    b /= 2;
    a = (a * a) % M;
  }
  return c;
}
ll fac[C], af[C], inv[C], pw[C];
void formulate_fac(int n) {
  int i;
  fac[0] = af[0] = 1;
  for (i = 1; i < n; i++) fac[i] = (i * fac[i - 1]) % M;
  for (i = 1; i < n; i++) af[i] = (pow(i, M - 2, M) * af[i - 1]) % M;
}
inline ll cn(int n, int k) {
  if (n < k || n < 0 || k < 0) return 0;
  return (((fac[n] * af[k]) % M) * af[n - k]) % M;
}
int main() {
  int n, i, j, sgn_outer, sgn_inner, k;
  ll res = 0, partial_res, summa = 0, l;
  formulate_fac(C);
  pw[0] = 1;
  for (i = 1; i < C; i++) inv[i] = pow(i, M - 2, M);
  for (i = 1; i < C; i++) pw[i] = (pw[i - 1] * 2) % M;
  scanf("%d %d %lld", &n, &k, &l);
  sgn_outer = -1;
  for (i = k; i <= n; i++) {
    sgn_outer *= (-1);
    partial_res = (((cn(n, i) * pw[i]) % M) * cn(i - 1, k - 1)) % M;
    summa = 0, sgn_inner = -1;
    for (j = 0; j <= i; j++) {
      sgn_inner *= (-1);
      summa = (summa + sgn_inner * cn(i, j) * inv[i + j + 1]) % M;
    }
    res = (res + partial_res * summa * sgn_outer) % M;
  }
  if (res < 0) res += M;
  printf("%lld\n", (l * res) % M);
  return 0;
}