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 = 2005, MOD = 998244353;
long long dp[MAXN * 2][MAXN], l, ans, f[MAXN];
int n, k;
long long rev(long long a) {
  int b = MOD - 2;
  long long res = 1;
  do {
    if (b % 2) res *= a, res %= MOD;
    a *= a, a %= MOD;
  } while (b /= 2);
  return res;
}
int main() {
  cin >> n >> k >> l;
  dp[0][0] = 1;
  for (int i = 1; i <= 2 * n; i++)
    for (int j = 0; j <= min(n, i); j++) {
      dp[i][j] += dp[i - 1][j + 1] * (j + 1) % MOD;
      if (j > 0) dp[i][j] += dp[i - 1][j - 1];
      dp[i][j] %= MOD;
    }
  f[0] = 1;
  for (int i = 1; i < MAXN; i++) f[i] = f[i - 1] * i % MOD;
  long long M = l * rev(dp[2 * n][0] * (2 * n + 1) % MOD) % MOD;
  for (int i = 1; i < 2 * n; i++)
    for (int j = k; j <= n; j++) {
      long long w = dp[i][j] * dp[2 * n - i][j] % MOD;
      w *= f[j], w %= MOD;
      ans += w * M % MOD;
      ans %= MOD;
    }
  cout << ans;
}