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 INF = 0x3f3f3f3f;
const int N = 4e3 + 10;
const long long mod = 998244353;
const bool debug = false;
int dp[N][N][2];
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int F[N], Finv[N], inv[N];
void init() {
  inv[1] = 1;
  for (int i = 2; i < N; i++) {
    inv[i] = (mod - mod / i) * 1ll * inv[mod % i] % mod;
  }
  F[0] = Finv[0] = 1;
  for (int i = 1; i < N; i++) {
    F[i] = F[i - 1] * 1ll * i % mod;
    Finv[i] = Finv[i - 1] * 1ll * inv[i] % mod;
  }
}
int main() {
  int n, k, l;
  scanf("%d%d%d", &n, &k, &l);
  init();
  dp[0][0][0] = 1;
  for (int i = 0; i <= 2 * n; ++i) {
    for (int j = 0; i + j <= 2 * n + 1 && j <= i; ++j) {
      if (j >= k) add(dp[i + 1][j][1], dp[i][j][0]);
      for (int z = 0; z <= 1; ++z) {
        add(dp[i + 1][j + 1][z], dp[i][j][z]);
        add(dp[i + 1][j - 1][z], dp[i][j][z] * 1ll * j % mod);
      }
    }
  }
  int ans = dp[2 * n + 1][0][1];
  for (int i = 1; i <= n; ++i) add(ans, ans);
  ans = ans * 1ll * F[n] % mod * Finv[2 * n + 1] % mod * l % mod;
  printf("%d\n", ans);
  return 0;
}