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 modpow(long long a, long long b, long long mod) {
  if (!b) return 1;
  a %= mod;
  return modpow(a * a % mod, b / 2, mod) * (b & 1 ? a : 1) % mod;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int DEBUG = 1;
const int mxn = 4007, M = 998244353;
int n, k, l;
long long dp[mxn][mxn][2];
void add(long long &a, long long b) {
  a += b;
  if (a >= M) a -= M;
}
void solve() {
  for (int i = 0; i < mxn; i++)
    for (int j = 0; j < mxn; j++)
      for (int k = 0; k < 2; k++) dp[i][j][k] = 0;
  dp[0][0][0] = 1;
  for (int i = 0; i <= 2 * n; i++)
    for (int j = 0; j <= n; j++)
      for (int x = 0; x <= 1; x++)
        if (dp[i][j][x]) {
          if (j)
            dp[i + 1][j - 1][x] = (dp[i + 1][j - 1][x] + j * dp[i][j][x]) % M;
          if (j < n)
            dp[i + 1][j + 1][x] = (dp[i + 1][j + 1][x] + dp[i][j][x]) % M;
          if (j >= k && !x)
            dp[i + 1][j][1] = (dp[i + 1][j][1] + dp[i][j][x]) % M;
        }
  long long ans = dp[2 * n + 1][0][1], deno = 1;
  for (int i = n + 1; i <= 2 * n + 1; i++) deno = deno * i % M;
  ans = ans * modpow(2, n, M) % M;
  ans = ans * modpow(deno, M - 2, M) % M * l % M;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  while (cin >> n >> k >> l) solve();
  return 0;
}