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;
const int N = 4010, mod = 998244353;
int n, k, l, f[N][N], fac[N], ans;
int ksm(int x, int y) {
  int s = 1;
  for (; y; y >>= 1, x = (long long)x * x % mod)
    if (y & 1) s = (long long)s * x % mod;
  return s;
}
inline void upd(int &x, int y) {
  x += y;
  x -= x >= mod ? mod : 0;
}
int main() {
  scanf("%d%d%d", &n, &k, &l);
  l = (long long)l * ksm(n << 1 | 1, mod - 2) % mod;
  fac[0] = 1;
  for (int i = (1); i <= (n); i++) fac[i] = (long long)fac[i - 1] * i % mod;
  f[0][0] = 1;
  int tmp;
  for (int i = (1); i <= (n << 1); i++)
    for (int j = (0); j <= (min(n, i - 1)); j++)
      if (tmp = f[i - 1][j]) {
        upd(f[i][j + 1], tmp);
        if (j) upd(f[i][j - 1], (long long)tmp * j % mod);
      }
  for (int i = (1); i <= (n << 1); i++)
    for (int j = (k); j <= (min(n, i)); j++)
      upd(ans, (long long)f[i][j] * f[(n << 1) - i][j] % mod * fac[j] % mod);
  ans = (long long)ans * ksm(f[n << 1][0], mod - 2) % mod * l % mod;
  printf("%d\n", ans);
  return 0;
}