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 N = 4005, P = 998244353;
using ll = long long;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % P;
    a = a * a % P;
    b >>= 1;
  }
  return ret;
}
inline void add(ll &a, ll b) { a = (a + b) % P; }
ll dp[N][N][2], fac[N];
int main() {
  ios::sync_with_stdio(false);
  int n, kk, l;
  cin >> n >> kk >> l;
  dp[0][0][0] = 1;
  for (int i = 0; i <= n * 2; i++)
    for (int j = 0; j <= i; j++)
      for (int k = 0; k < 2; k++) {
        add(dp[i + 1][j + 1][k], dp[i][j][k]);
        if (!k && j >= kk) add(dp[i + 1][j][1], dp[i][j][k]);
        if (j) add(dp[i + 1][j - 1][k], dp[i][j][k] * j % P);
      }
  fac[0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++) fac[i] = fac[i - 1] * i % P;
  cout << dp[n * 2 + 1][0][1] * qpow(2, n) % P * fac[n] % P *
              qpow(fac[n * 2 + 1], P - 2) % P * l % P;
  return 0;
}