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 mod = 998244353;
long long n, k, l;
vector<long long> fa, in;
vector<vector<long long> > dp;
long long po(long long x, long long p) {
  long long res = 1;
  while (p) {
    if (p % 2) {
      res = (res * x) % mod;
      --p;
    } else {
      x = (x * x) % mod;
      p /= 2;
    }
  }
  return res;
}
long long c(long long n, long long k) {
  return (fa[n] * ((in[k] * in[n - k]) % mod)) % mod;
}
long long add(long long& x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
long long mul(long long x, long long y) { return (x * y) % mod; }
long long mul(long long x, long long y, long long z) {
  return mul(x, mul(y, z));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k >> l;
  fa.assign(5000, 1);
  in.assign(5000, 1);
  for (int i = 2; i < 5000; ++i) {
    fa[i] = (fa[i - 1] * i) % mod;
    in[i] = po(fa[i], mod - 2);
  }
  long long q = mul(2 * n + 1, in[n]);
  for (int i = 2 * n; i > 2; i -= 2) q = mul(q, c(i, 2));
  q = po(q, mod - 2);
  dp.assign(n + 1, vector<long long>(n + 1, 0));
  dp[0][0] = 1;
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) {
      if (dp[i][j] == 0) continue;
      if (i < n) add(dp[i + 1][j], dp[i][j]);
      if (j < i) add(dp[i][j + 1], mul(dp[i][j], i - j));
    }
  }
  long long p = 0;
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) {
      if (i - j < k) continue;
      add(p, mul(fa[i - j], dp[i][j], dp[n - j][n - i]));
    }
  }
  cout << mul(l, p, q);
  return 0;
}