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;
int factorial[2010];
int choose[2010][2010];
int inverseCache[2010 * 2];
int poly[2010 * 2];
int temp[2010 * 2];
int realpoly[2010 * 2];
int exclIncl[2010];
inline int add(int a, int b) { return (a + b) % 998244353; }
inline int prod(int a, int b) { return (long long)a * b % 998244353; }
inline int sub(int a, int b) { return (998244353 + a - b) % 998244353; }
int pow(int a, int n) {
  if (!n) return 1;
  int half = pow(a, n / 2);
  half = prod(half, half);
  if (n % 2) half = prod(half, a);
  return half;
}
int inverse(int a) { return pow(a, 998244353 - 2); }
void init() {
  factorial[0] = 1;
  for (int i = 1; i < 2010; i++) factorial[i] = prod(factorial[i - 1], i);
  for (int i = 0; i < 2010 * 2; i++) inverseCache[i] = inverse(i);
  choose[0][0] = 1;
  for (int i = 0; i < 2010; i++) {
    choose[i][0] = 1;
    for (int j = 1; j < i; j++)
      choose[i][j] = add(choose[i - 1][j - 1], choose[i - 1][j]);
    choose[i][i] = 1;
  }
}
int main(void) {
  int N, K, L;
  cin >> N >> K >> L;
  init();
  int ans = factorial[N];
  poly[0] = 1;
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < 2 * 2010 - 2; j++) {
      temp[j + 2] = sub(temp[j + 2], poly[j]);
      temp[j + 2] = sub(temp[j + 2], poly[j]);
      temp[j + 1] = add(temp[j + 1], poly[j]);
      temp[j + 1] = add(temp[j + 1], poly[j]);
    }
    for (int j = 0; j < 2 * 2010; j++) {
      poly[j] = temp[j];
      temp[j] = 0;
    }
  }
  for (int i = K; i <= N; i++) {
    int coeff = inverse(prod(factorial[i], factorial[N - i]));
    exclIncl[i] = 1;
    for (int j = K; j < i; j++)
      exclIncl[i] = sub(exclIncl[i], prod(choose[i][j], exclIncl[j]));
    for (int j = 0; j < 2 * 2010; j++)
      realpoly[j] = add(realpoly[j], prod(poly[j], prod(coeff, exclIncl[i])));
    for (int j = 0; j < 2 * 2010 - 2; j++) {
      temp[j + 2] = sub(temp[j + 2], poly[j]);
      temp[j + 2] = sub(temp[j + 2], poly[j]);
      temp[j + 1] = add(temp[j + 1], poly[j]);
      temp[j + 1] = add(temp[j + 1], poly[j]);
    }
    for (int j = 0; j < 2 * 2010; j++) {
      poly[j] = temp[j];
      temp[j] = 0;
    }
  }
  int integral = 0;
  for (int i = 0; i < 2 * 2010 - 1; i++)
    integral = add(integral, prod(realpoly[i], inverseCache[i + 1]));
  ans = prod(ans, integral);
  ans = prod(ans, L);
  cout << ans << endl;
  return 0;
}