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>
const int MAXN = 2010;
const int mod = 998244353;
void reduce(int& x) { x += x >> 31 & mod; }
int mul(int a, int b) { return (long long)a * b % mod; }
int pow(int a, int b, int res = 1) {
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) res = mul(res, a);
  return res;
}
int remod(long long x) {
  x %= mod;
  return x + (x >> 63 & mod);
}
int n, K, L, dp[MAXN * 2][MAXN];
int fac[MAXN];
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  for (int i = *fac = 1; i != MAXN; ++i) fac[i] = mul(fac[i - 1], i);
  std::cin >> n >> K >> L;
  dp[0][0] = 1;
  for (int i = 1; i <= n * 2; ++i) {
    for (int j = 0; j <= n; ++j) {
      if (j) reduce(dp[i][j] += dp[i - 1][j - 1] - mod);
      reduce(dp[i][j] += mul(dp[i - 1][j + 1], j + 1) - mod);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n * 2; ++i)
    for (int j = K; j <= n; ++j)
      reduce(ans +=
             (long long)dp[i][j] * dp[n * 2 - i][j] % mod * fac[j] % mod - mod);
  ans = pow(dp[n * 2][0], mod - 2, ans);
  ans = pow(2 * n + 1, mod - 2, mul(ans, L));
  std::cout << ans << std::endl;
  return 0;
}