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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
constexpr auto MOD = 998244353LL;
constexpr auto LINF = (1LL << 60);
constexpr auto INF = 2147483647LL;
constexpr auto PI = 3.1415926535897932384626433;
constexpr auto EPS = 1E-9;
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &x : v) in >> x;
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> v) {
  for (int i = 0; i < v.size(); i++)
    out << v[i] << (i + 1 == v.size() ? '\n' : ' ');
  out.flush();
  return out;
}
ll dp[2010][2010][2], len;
int n, k;
ll f[4010], c[4010];
ll nck(ll n, ll k) {
  if (n < k) return 0;
  ll ret = (f[n] * c[k]) % MOD;
  return (ret * c[n - k]) % MOD;
}
ll npk(ll n, ll k) {
  if (n < k) return 0;
  return f[n] * c[n - k] % MOD;
}
ll pwr(ll b, ll e) {
  b %= MOD;
  if (e == 0) return 1;
  if (e == 1) return b;
  return (pwr(b * b, e / 2) * pwr(b, e % 2)) % MOD;
}
void init() {
  f[0] = c[0] = 1;
  for (int i = 1; i < 4010; i++) (f[i] = f[i - 1] * i) %= MOD;
  c[4010 - 1] = pwr(f[4010 - 1], MOD - 2);
  for (int i = 4010 - 2; i; i--) (c[i] = c[i + 1] * (i + 1)) %= MOD;
  return;
}
ll dpfs(int x, int y, int f) {
  if (x > n) return 0;
  if (y > x) return 0;
  if (x == n && y == n && f == 1) return 1;
  ll &res = dp[x][y][f];
  if (res != -1) return res;
  res = 0;
  if (f == 0) {
    if (x - y >= k) {
      res += dpfs(x, y, 1);
    }
  }
  if (x < n) {
    res += dpfs(x + 1, y, f);
  }
  if (x > y) {
    res += (x - y) * dpfs(x, y + 1, f);
  }
  res %= MOD;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  init();
  cin >> n >> k >> len;
  memset(dp, -1, sizeof(dp));
  cout << len * dpfs(0, 0, 0) % MOD * f[n] % MOD * pwr(2, n) % MOD *
              pwr(f[2 * n + 1], MOD - 2) % MOD
       << endl;
  return 0;
}