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 = 4010, mod = 998244353;
int n, k, l, ni[maxn], nijie[maxn], ans, jie[maxn], f[maxn][maxn][2];
int qpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1LL * res * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  scanf("%d%d%d", &n, &k, &l);
  ni[0] = ni[1] = nijie[0] = nijie[1] = jie[0] = jie[1] = 1;
  for (int i = 2; i < maxn; i++) {
    jie[i] = 1LL * jie[i - 1] * i % mod;
    ni[i] = 1LL * (mod - mod / i) * ni[mod % i] % mod;
    nijie[i] = 1LL * nijie[i - 1] * ni[i] % mod;
  }
  f[0][0][0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++)
    for (int j = 0; j < i; j++) {
      (f[i][j + 1][0] += f[i - 1][j][0]) %= mod;
      (f[i][j + 1][1] += f[i - 1][j][1]) %= mod;
      if (j) (f[i][j - 1][0] += 1LL * f[i - 1][j][0] * j % mod) %= mod;
      if (j) (f[i][j - 1][1] += 1LL * f[i - 1][j][1] * j % mod) %= mod;
      if (j >= k) (f[i][j][1] += f[i - 1][j][0]) %= mod;
    }
  ans = f[2 * n + 1][0][1];
  ans = 1LL * ans * l % mod;
  ans = 1LL * ans * qpow(2, n) % mod;
  ans = 1LL * ans * jie[n] % mod;
  ans = 1LL * ans * nijie[2 * n + 1] % mod;
  printf("%d\n", ans);
  return 0;
}