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 uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &item) {
  out << '(' << item.first << ", " << item.second << ')';
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (const auto &item : v) out << item << ' ';
  return out;
}
const int NMAX = 4010;
const int KMAX = 2010;
const int MOD = 998244353;
int dp[NMAX][KMAX][2];
void add(int &x, int y) { x = (x + y < MOD ? x + y : x + y - MOD); }
int pw(int base, int exp) {
  int res;
  for (res = 1; exp; exp >>= 1) {
    if (exp & 1) res = 1LL * res * base % MOD;
    base = 1LL * base * base % MOD;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, x, n, k, l, fact, ans;
  cin >> n >> k >> l;
  dp[0][0][0] = 1;
  for (i = 0; i < 2 * n + 1; ++i)
    for (j = 0; j <= i && j <= n; ++j)
      for (x = 0; x <= 1; ++x) {
        if (!dp[i][j][x]) continue;
        if (j + 1 <= 2 * n - (i + 1) + x) add(dp[i + 1][j + 1][x], dp[i][j][x]);
        if (j) add(dp[i + 1][j - 1][x], 1LL * dp[i][j][x] * j % MOD);
        if (j >= k && x == 0) add(dp[i + 1][j][1], dp[i][j][x]);
      }
  for (fact = 1, i = n + 1; i <= 2 * n + 1; ++i) fact = 1LL * fact * i % MOD;
  ans = dp[2 * n + 1][0][1];
  ans = 1LL * ans * l % MOD;
  ans = 1LL * ans * pw(fact, MOD - 2) % MOD;
  ans = 1LL * ans * pw(2, n) % MOD;
  cout << ans << '\n';
  return 0;
}