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>
int ch[1 + 2000][1 + 2000];
void init(int n) {
  int i, j;
  ch[0][0] = 1;
  for (i = 1; i <= n; i++) {
    ch[i][0] = 1;
    for (j = 1; j <= i; j++)
      ch[i][j] = (ch[i - 1][j] + ch[i - 1][j - 1]) % 998244353;
  }
}
long long power(int a, int k) {
  long long p;
  if (k == 0) return 1;
  p = power(a, k / 2);
  p = p * p % 998244353;
  if (k % 2) p = p * a % 998244353;
  return p;
}
int main() {
  static long long pp[1 + 2000 + 2000], qq[1 + 2000 + 2000],
      aa[1 + 2000 + 2000];
  int n, k, l, i, j;
  long long inv2, a;
  scanf("%d%d%d", &n, &k, &l);
  init(n);
  pp[n] = power(2, n);
  for (i = 0; i < n; i++)
    for (j = n + n; j > 0; j--) pp[j] = (pp[j] - pp[j - 1]) % 998244353;
  for (j = 0; j <= n + n; j++) aa[j] = pp[j];
  inv2 = (998244353 + 1) / 2;
  for (i = 1; i <= n - k; i++) {
    for (j = 1; j < n + n; j++) qq[j] = pp[j + 1] * inv2 % 998244353;
    qq[n + n] = 0;
    for (j = 1; j <= n + n; j++) qq[j] = (qq[j] + qq[j - 1]) % 998244353;
    for (j = 0; j <= n + n; j++) pp[j] = (qq[j] - pp[j]) % 998244353;
    a = ch[n][i];
    for (j = 0; j <= n + n; j++) aa[j] = (aa[j] + a * pp[j]) % 998244353;
  }
  a = 0;
  for (i = 0; i <= n + n; i++)
    a = (a + aa[i] * power(i + 1, 998244353 - 2)) % 998244353;
  a = (a * l % 998244353 + 998244353) % 998244353;
  printf("%lld\n", a);
  return 0;
}