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;
int fact[4005];
int dp[4005][4005];
int powi(int a, int b) {
  if (!b) return 1;
  int p = powi(a, b / 2);
  p = (1LL * p * p) % 998244353;
  if (b & 1) p = (1LL * p * a) % 998244353;
  return p;
}
int mmi(int x) { return powi(x, 998244353 - 2); }
int main() {
  fact[0] = 1;
  for (int i = 1; i < 4005; i++) fact[i] = 1LL * fact[i - 1] * i % 998244353;
  int n, k, l;
  scanf("%d %d %d", &n, &k, &l);
  n *= 2;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i][j] = 1LL * (j + 1) * dp[i - 1][j + 1] % 998244353;
      if (j) dp[i][j] += dp[i - 1][j - 1], dp[i][j] %= 998244353;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = k; j <= i; j++) {
      int tmp = 1LL * dp[i][j] * dp[n - i][j] % 998244353;
      tmp = 1LL * tmp * fact[j] % 998244353;
      ans += tmp, ans %= 998244353;
    }
  }
  ans = 1LL * ans * mmi(dp[n][0]) % 998244353;
  ans = 1LL * ans * l % 998244353;
  ans = 1LL * ans * mmi(n + 1) % 998244353;
  printf("%d\n", ans);
  return 0;
}