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;
int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
int mul(int a, int b) { return (long long)a * b % MOD; }
int qpow(int a, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) ans = mul(ans, a);
    a = mul(a, a);
    p >>= 1;
  }
  return ans;
}
int inv(int a) { return qpow(a, MOD - 2); }
int divv(int a, int b) { return mul(a, inv(b)); }
const int MAXN = 2009 << 1;
int fact[MAXN], invFact[MAXN], pow2[MAXN], N, k, l;
int nCk(int a, int b) { return mul(fact[a], mul(invFact[a - b], invFact[b])); }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (int i = 1; i <= MAXN - 1; ++i) fact[i] = mul(fact[i - 1], i);
  invFact[MAXN - 1] = inv(fact[MAXN - 1]);
  for (int i = MAXN - 1; i >= 1; --i) invFact[i - 1] = mul(invFact[i], i);
  pow2[0] = 1;
  for (int i = 1; i <= MAXN - 1; ++i) pow2[i] = mul(pow2[i - 1], 2);
  cin >> N >> k >> l;
  int ans = 0;
  for (int i = 0; i <= 2 * N; ++i)
    for (int j = i; j >= k; j -= 2) {
      if (i + j > 2 * N) continue;
      int x = 1;
      x = mul(x, nCk(i, j));
      x = mul(x, nCk(2 * N - i, j));
      x = mul(x, fact[i - j]);
      x = mul(x, fact[2 * N - j - i]);
      x = mul(x, fact[j]);
      x = mul(x, pow2[j]);
      x = mul(x, invFact[(i - j) / 2]);
      x = mul(x, invFact[(2 * N - j - i) / 2]);
      ans = add(ans, x);
    }
  ans = mul(ans, fact[N]);
  ans = divv(ans, fact[2 * N]);
  ans = divv(ans, 2 * N + 1);
  ans = mul(ans, l % MOD);
  cout << ans << endl;
  return 0;
}