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 N = 4010, mod = 998244353;
int ksm(long long a, int b, int c = 1) {
  for (; b; b /= 2, a = a * a % mod)
    if (b & 1) c = c * a % mod;
  return c;
}
int n, K, L, dp[N][N], ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> K >> L;
  dp[0][0] = 1;
  for (int i = 1; i <= 2 * n; ++i)
    for (int j = 0; j <= n; ++j)
      if (dp[i - 1][j]) {
        dp[i][j + 1] = (dp[i][j + 1] + dp[i - 1][j]) % mod;
        if (j)
          dp[i][j - 1] = (dp[i][j - 1] + (long long)dp[i - 1][j] * j) % mod;
      }
  for (int i = 1; i <= 2 * n; ++i)
    for (int j = 1, k = 1; j <= n; ++j, k = (long long)k * j % mod)
      if (j >= K)
        ans = (ans + (long long)dp[i][j] * dp[n * 2 - i][j] % mod * k) % mod;
  cout << ksm((long long)(2 * n + 1) * dp[2 * n][0] % mod, mod - 2,
              (long long)ans * L % mod)
       << '\n';
  return 0;
}