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;
long long dp[2100 << 1][2100][2];
long long n, k, l;
long long mypow(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * x % 998244353;
    n /= 2;
    x = x * x % 998244353;
  }
  return ans % 998244353;
}
int main() {
  cin >> n >> k >> l;
  dp[0][0][0] = 1;
  for (register int i = 1; i < 2 * n + 2; i++)
    for (register int j = 0; j < min((long long)i, n) + 1; j++) {
      if (j >= k) dp[i][j][1] += dp[i - 1][j][0];
      for (register int x = 0; x < 2; x++) {
        if (i + j - x <= 2 * n && j > 0) dp[i][j][x] += dp[i - 1][j - 1][x];
        dp[i][j][x] += dp[i - 1][j + 1][x] * (j + 1);
        dp[i][j][x] %= 998244353;
      }
    }
  long long tem1, tem2;
  tem1 = 1;
  for (register int i = 1; i < n + 1; i++) tem1 = tem1 * 2 * i % 998244353;
  tem2 = 1;
  for (register int i = 1; i < 2 * n + 2; i++) tem2 = tem2 * i % 998244353;
  long long ans = l;
  ans = ans * dp[2 * n + 1][0][1] % 998244353;
  ans = ans * tem1 % 998244353;
  ans = ans * mypow(tem2, 998244353 - 2) % 998244353;
  printf("%lld", ans);
  return ~~(0 ^ 0 ^ 0);
}