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 MAXN = 2e3;
const int MOD = 998244353;
int n, k, l;
long long inverse[2 * MAXN + 2], factorial[2 * MAXN + 2],
    inversefactorial[2 * MAXN + 2], power2[2 * MAXN + 2], dp[MAXN + 1];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> k >> l;
  inverse[1] = 1;
  for (int i = 2; i < 2 * MAXN + 2; ++i)
    inverse[i] = MOD - MOD / i * inverse[MOD % i] % MOD;
  factorial[0] = inversefactorial[0] = power2[0] = 1;
  for (int i = 1; i < 2 * MAXN + 2; ++i) {
    factorial[i] = factorial[i - 1] * i % MOD;
    inversefactorial[i] = inversefactorial[i - 1] * inverse[i] % MOD;
    power2[i] = power2[i - 1] * 2 % MOD;
  }
  for (int i = n; i >= k; --i) {
    dp[i] = factorial[i] * factorial[i] % MOD * power2[i] % MOD *
            inversefactorial[2 * i + 1] % MOD;
    dp[i] = dp[i] * factorial[n] % MOD * inversefactorial[i] % MOD *
            inversefactorial[n - i] % MOD;
    for (int j = i + 1; j <= n; ++j) {
      dp[i] = (dp[i] -
               factorial[j] * inversefactorial[i] % MOD *
                   inversefactorial[j - i] % MOD * dp[j] % MOD +
               MOD) %
              MOD;
    }
  }
  long long ans = 0;
  for (int i = k; i <= n; ++i) ans = (ans + dp[i]) % MOD;
  cout << ans * l % MOD << endl;
  return 0;
}