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;
inline int read() {
  int out = 0, fh = 1;
  char jp = getchar();
  while ((jp > '9' || jp < '0') && jp != '-') jp = getchar();
  if (jp == '-') fh = -1, jp = getchar();
  while (jp >= '0' && jp <= '9') out = out * 10 + jp - '0', jp = getchar();
  return out * fh;
}
const int P = 998244353;
inline int add(int a, int b) { return (a + b) % P; }
inline int mul(int a, int b) { return 1LL * a * b % P; }
int fpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int inv(int x) { return fpow(x, P - 2); }
const int MAXN = 2019 << 1;
int n, l, k;
int fac[MAXN];
int f[MAXN][MAXN];
int main() {
  n = read(), k = read();
  l = read();
  int perl = mul(l, inv(2 * n + 1));
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = mul(fac[i - 1], i);
  f[0][0] = 1;
  for (int i = 0; i < 2 * n; ++i) {
    for (int j = n < i ? n : i; j >= 0; --j) {
      f[i + 1][j + 1] = add(f[i + 1][j + 1], f[i][j]);
      if (j) f[i + 1][j - 1] = add(f[i + 1][j - 1], mul(f[i][j], j));
    }
  }
  int ans = 0;
  for (int i = 1; i < 2 * n; ++i)
    for (int j = k; j <= n; ++j) {
      int tmp = mul(f[i][j], f[2 * n - i][j]);
      tmp = mul(tmp, fac[j]);
      ans = add(ans, tmp);
    }
  cout << mul(ans, mul(perl, inv(f[2 * n][0])));
  return 0;
}