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[2 * 2001][2001][2];
long long ModularExponentation(long long a, long long n) {
  long long ans = 1 % 998244353;
  for (; n; n /= 2) {
    if (n % 2) {
      ans = (ans * a) % 998244353;
    }
    a = (a * a) % 998244353;
  }
  return ans;
}
long long ModularInverse(long long a) {
  return ModularExponentation(a, 998244353 - 2);
}
long long Fact(int n) {
  long long ans = 1;
  for (int i = 1; i <= n; ++i) {
    ans = (ans * i) % 998244353;
  }
  return ans;
}
void UpdateDp(long long &dpVal, long long val) {
  dpVal = (dpVal + val) % 998244353;
}
int main() {
  int n, k, l;
  cin >> n >> k >> l;
  dp[0][0][0] = 1;
  for (int i = 1; i <= 2 * n + 1; ++i) {
    for (int j = 0, m = min(i, n); j <= m; ++j) {
      if (j >= k) {
        UpdateDp(dp[i][j][1], dp[i - 1][j][0]);
      }
      if (j) {
        UpdateDp(dp[i][j][0], dp[i - 1][j - 1][0]);
        UpdateDp(dp[i][j][1], dp[i - 1][j - 1][1]);
      }
      if (j <= i - 2) {
        UpdateDp(dp[i][j][0], (j + 1) * dp[i - 1][j + 1][0]);
        UpdateDp(dp[i][j][1], (j + 1) * dp[i - 1][j + 1][1]);
      }
    }
  }
  long long ans = dp[2 * n + 1][0][1];
  for (int i = 1; i <= n; ++i) {
    ans = (ans * 2 * i) % 998244353;
  }
  ans = (ans * ModularInverse(Fact(2 * n + 1))) % 998244353;
  ans = (l * ans) % 998244353;
  cout << ans << '\n';
  return 0;
}