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 mod = 998244353;
long long f[4005], g[4005], p[4005];
long long cal(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y % 2) ret = ret * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ret;
}
long long comb(long long x, long long y) {
  long long num = f[x];
  long long denum = g[y] * g[x - y] % mod;
  return num * denum % mod;
}
long long w(long long x) {
  long long ret = f[x] * f[x] % mod;
  ret = ret * g[2 * x + 1] % mod;
  return ret;
}
int main() {
  f[0] = g[0] = p[0] = 1;
  for (int i = 1; i <= 4004; i++) {
    f[i] = f[i - 1] * i % mod;
    g[i] = cal(f[i], mod - 2);
    p[i] = p[i - 1] * 2 % mod;
  }
  int n, k, l;
  cin >> n >> k >> l;
  long long ans = 0;
  for (int i = k; i <= n; i++) {
    long long temp = 0;
    for (int j = 0; j <= n - i; j++) {
      long long tmp = comb(n - i, j) * p[i + j] % mod;
      if (j % 2) tmp = -tmp;
      tmp = tmp * w(i + j) % mod;
      temp += tmp;
    }
    temp %= mod;
    temp = temp * comb(n, i) % mod;
    ans += temp;
  }
  ans %= mod;
  ans = ans * l % mod;
  ans += mod;
  ans %= mod;
  cout << ans << endl;
}