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 int INF = 4096;
long long dp[2][INF][2];
long long rev(long long a) {
  long long ans = 1;
  int b = mod - 2;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k, l;
  cin >> n >> k >> l;
  dp[0][0][0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k < 2; k++) {
        if (j > 0) dp[1][j][k] = (dp[1][j][k] + dp[0][j - 1][k]) % mod;
        dp[1][j][k] = (dp[0][j + 1][k] * (j + 1) % mod + dp[1][j][k]) % mod;
      }
      if (j >= k) dp[1][j][1] = (dp[0][j][0] + dp[1][j][1]) % mod;
    }
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k < 2; k++) dp[0][j][k] = dp[1][j][k], dp[1][j][k] = 0;
    }
  }
  long long ans = dp[0][0][1];
  for (int i = n; i >= 1; i--) ans = ans * i % mod;
  for (int i = 0; i < n; i++) ans = ans * 2 % mod;
  long long base = 1;
  for (int i = 1; i <= 2 * n + 1; i++) base = base * i % mod;
  ans = ans * rev(base) % mod * l % mod;
  cout << ans << endl;
}