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 long long mod = 998244353;
const long long maxn = 2100;
long long n, L, k;
long long prob[maxn * 2];
long long pow2[maxn * 2], rev2[maxn * 2];
long long C[maxn * 2][maxn * 2], fact[maxn * 2];
long long g[maxn * 2];
long long tmp, ans;
long long get(long long x) { return g[x]; }
long long poww(long long f, long long x) {
  long long s = 1;
  while (x) {
    if (x % 2) s = (s * f) % mod;
    f = (f * f) % mod;
    x = x / 2;
  }
  return s;
}
long long inv(long long x) { return poww(x, mod - 2); }
void work() {
  long long i, j, l, r;
  scanf("%lld%lld%lld", &n, &k, &L);
  pow2[0] = 1;
  rev2[0] = 1;
  fact[0] = 1;
  for (i = 1; i <= n * 2 + 2; i++) {
    pow2[i] = (pow2[i - 1] * 2) % mod;
    rev2[i] = inv(pow2[i]);
    fact[i] = (fact[i - 1] * i) % mod;
  }
  C[0][0] = 1;
  for (i = 1; i <= n * 2 + 2; i++) {
    C[i][0] = 1;
    for (j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
  g[0] = 1;
  for (i = 2; i <= n * 2 + 2; i += 2) {
    g[i] = (g[i - 2] * (i - 1)) % mod;
  }
  for (i = 1; i <= n * 2; i++) {
    prob[i] = 0;
    for (j = k; j <= n * 2; j++) {
      l = i;
      r = 2 * n - i;
      if (l < j || r < j || (l - j) % 2 != 0) continue;
      tmp = 0;
      tmp = (C[l][j] * C[r][j]) % mod;
      tmp = (tmp * fact[j]) % mod;
      tmp = (tmp * get(l - j)) % mod;
      tmp = (tmp * get(r - j)) % mod;
      prob[i] = (prob[i] + tmp) % mod;
      ans = (ans + tmp) % mod;
    }
  }
  ans = (ans * inv(n * 2 + 1)) % mod;
  ans = (ans * inv(get(n * 2))) % mod;
  ans = (ans * L) % mod;
  printf("%lld\n", ans);
}
int main() {
  work();
  return 0;
}