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 maxn = 2010;
const long long mod = 998244353;
long long ksm(long long a, long long k) {
  long long ret = 1ll, x = a;
  while (k) {
    if (k & 1) ret = ret * x % mod;
    k >>= 1;
    x = x * x % mod;
  }
  return ret;
}
long long n, k, l, f[maxn * 2][maxn][2], ans1, ans2;
int main() {
  scanf("%lld%lld%lld", &n, &k, &l);
  f[0][0][0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++)
    for (int j = 0; j <= n; j++) {
      if (j >= k) f[i][j][1] += f[i - 1][j][0], f[i][j][1] %= mod;
      for (int x = 0; x <= 1; x++) {
        if (i + j + 1 - x <= 2 * n + 1 && j - 1 >= 0)
          f[i][j][x] += f[i - 1][j - 1][x], f[i][j][x] %= mod;
        f[i][j][x] += (j + 1) * f[i - 1][j + 1][x] % mod, f[i][j][x] %= mod;
      }
    }
  ans1 = f[2 * n + 1][0][1];
  ans2 = 1;
  for (int i = 1; i <= n; i++) ans1 = ans1 * i % mod, ans1 = ans1 * 2 % mod;
  for (int i = 1; i <= 2 * n + 1; i++) ans2 = ans2 * i % mod;
  printf("%lld\n", ans1 * ksm(ans2, mod - 2) % mod * l % mod);
  return 0;
}