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 M = 4005;
int dp[M][M][2];
int mod = 998244353;
const int N = 2e6 + 100;
namespace Combi {
int fact[N], inv[N];
int bm(int b, int p, int m) {
  if (p == 0) return 1 % m;
  int t = bm(b, p / 2, m);
  t = (1ll * t * t) % m;
  if (p & 1) return 1ll * t * b % m;
  return t;
}
int C(int n, int r) {
  if (n < 0 or r < 0 or r > n) return 0;
  int ret = 1ll * fact[n] * inv[r] % mod;
  ret = 1ll * ret * inv[n - r] % mod;
  return ret;
}
int no_of_eqns(int var, int sum) { return C(sum + var - 1, var - 1); }
void init() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = 1ll * fact[i - 1] * i % mod;
  }
  inv[N - 1] = bm(fact[N - 1], mod - 2, mod);
  for (int i = N - 2; i >= 0; i--) {
    inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  }
}
}  // namespace Combi
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  Combi::init();
  int n, k, l;
  cin >> n >> k >> l;
  dp[0][0][0] = 1;
  for (int i = 0; i < 2 * n + 1; i++) {
    for (int unmatched = 0; unmatched <= i; unmatched++) {
      if (unmatched >= k) add(dp[i + 1][unmatched][1], dp[i][unmatched][0]);
      if (unmatched)
        add(dp[i + 1][unmatched - 1][0],
            1ll * dp[i][unmatched][0] * unmatched % mod);
      if (unmatched)
        add(dp[i + 1][unmatched - 1][1],
            1ll * dp[i][unmatched][1] * unmatched % mod);
      add(dp[i + 1][unmatched + 1][0], dp[i][unmatched][0]);
      add(dp[i + 1][unmatched + 1][1], dp[i][unmatched][1]);
    }
  }
  int ans = dp[2 * n + 1][0][1];
  ans = 1ll * ans * Combi::fact[n] % mod;
  ans = 1ll * ans * Combi::bm(2, n, mod) % mod;
  ans = 1ll * ans * Combi::inv[2 * n + 1] % mod;
  ans = 1ll * ans * l % mod;
  cout << ans << endl;
  return 0;
}