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 = 2e3 + 10, mod = 998244353;
long long choose[N][N], pre_inverse[2 * N];
long long pwr(long long x, long long e);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  long long l;
  cin >> n >> k >> l;
  for (int i = 0; i < N; ++i) {
    choose[i][0] = choose[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      choose[i][j] = (choose[i - 1][j - 1] + choose[i - 1][j]) % mod;
    }
  }
  for (int i = 1; i < 2 * N; ++i) {
    pre_inverse[i] = pwr(i, mod - 2);
  }
  long long ans = 0, p2 = 1;
  for (int j = 1; j <= n; ++j) {
    p2 *= 2;
    p2 %= mod;
    long long sum = 0;
    for (int i = k; i <= j; ++i) {
      long long temp = choose[n][i] * choose[n - i][j - i];
      temp %= mod;
      if ((j - i) % 2) {
        temp = (mod - temp);
      }
      sum += temp;
      sum %= mod;
    }
    for (int t = 0; t <= j; ++t) {
      long long temp = choose[j][t];
      temp *= p2;
      temp %= mod;
      temp *= l;
      temp %= mod;
      temp *= pre_inverse[2 * j - t + 1];
      temp %= mod;
      temp *= sum;
      temp %= mod;
      if ((j - t) % 2) {
        temp = (mod - temp);
      }
      ans += temp;
      ans %= mod;
    }
  }
  cout << ans << endl;
}
long long pwr(long long x, long long e) {
  long long res = 1;
  while (e) {
    if (e & 1) {
      res = (res * x) % mod;
    }
    e /= 2;
    x = (x * x) % mod;
  }
  return res;
}