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>
const int P = 998244353;
inline int mul(const int &a, const int &b) { return 1ll * a * b % P; }
inline int add(int a, const int &b) {
  a += b;
  return (a >= P) ? a - P : a;
}
inline int sub(int a, const int &b) {
  a -= b;
  return (a < 0) ? a + P : a;
}
inline int inv(int a) { return (a <= 1) ? 1 : mul(P - P / a, inv(P % a)); }
int qsm(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
int f[4010][2010][2], n, k, l;
int main() {
  scanf("%d%d%d", &n, &k, &l);
  f[0][0][0] = 1;
  for (int i = 1; i <= (n << 1) + 1; i++)
    for (int j = 0; j <= n && j <= i; j++) {
      if (j >= k) f[i][j][1] = add(f[i][j][1], f[i - 1][j][0]);
      f[i][j][1] = add(f[i][j][1], f[i - 1][j - 1][1]);
      f[i][j][0] = add(f[i][j][0], f[i - 1][j - 1][0]);
      f[i][j][1] = add(f[i][j][1], mul(f[i - 1][j + 1][1], j + 1));
      f[i][j][0] = add(f[i][j][0], mul(f[i - 1][j + 1][0], j + 1));
    }
  int ans = f[(n << 1) + 1][0][1];
  for (int i = 1; i <= n + 1; i++) ans = mul(ans, inv(i + n));
  printf("%d\n", mul(mul(ans, l), qsm(2, n)));
}