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 N = 5010;
const int mod = 998244353;
int inf = 1000000000;
int n, k, L;
long long dp[N][N];
long long p[N];
long long pw(long long a, int b) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1) ret = (1ll * ret * a) % mod;
    a = (1ll * a * a) % mod;
    b = b >> 1;
  }
  return ret;
}
long long add(long long a, long long b) { return (a + b) % mod; }
long long sol() {
  p[0] = 1;
  for (int i = 1; i <= n; ++i) {
    p[i] = (p[i - 1] * i) % mod;
  }
  dp[1][1] = 1;
  for (int i = 1; i <= 2 * n; ++i) {
    for (int j = 0; j <= i; ++j) {
      if (i + j + 2 <= 2 * n)
        dp[i + 1][j + 1] = add(dp[i + 1][j + 1], dp[i][j]);
      if (j) dp[i + 1][j - 1] = add(dp[i + 1][j - 1], (dp[i][j] * j) % mod);
    }
  }
  long long ans = 0;
  for (int i = 1; i <= 2 * n; ++i) {
    for (int j = k; j <= n; ++j) {
      if (i + j > 2 * n) break;
      ans = (ans + (dp[i][j] * dp[2 * n - i][j]) % mod * p[j]) % mod;
    }
  }
  ans = (ans * pw(dp[2 * n][0], mod - 2)) % mod;
  ans = (ans * (L * pw(2 * n + 1, mod - 2) % mod)) % mod;
  return ans;
}
int main() {
  cin >> n >> k >> L;
  cout << sol() << endl;
}