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;
int n, k;
long long l;
long long mo = 998244353;
long long A[4005][4005][2];
long long B[4005];
long long a, b;
long long f(long long x, long long y) {
  long long mid = 1;
  while (y) {
    if (y & 1) {
      mid = mid * x % mo;
    }
    x = x * x % mo;
    y >>= 1;
  }
  return mid;
}
int main() {
  cin >> n >> k >> l;
  A[0][0][0] = 1;
  for (int i = 0; i <= 2 * n; i++) {
    for (int j = 0; j <= i; j++) {
      A[i + 1][j + 1][0] = (A[i + 1][j + 1][0] + A[i][j][0]) % mo;
      A[i + 1][j + 1][1] = (A[i + 1][j + 1][1] + A[i][j][1]) % mo;
      if (j) {
        A[i + 1][j - 1][0] = (A[i + 1][j - 1][0] + A[i][j][0] * j) % mo;
        A[i + 1][j - 1][1] = (A[i + 1][j - 1][1] + A[i][j][1] * j) % mo;
      }
      if (j >= k) {
        A[i + 1][j][1] = (A[i + 1][j][1] + A[i][j][0]) % mo;
      }
    }
  }
  B[0] = 1;
  for (int i = 1; i <= n * 2 + 1; i++) {
    B[i] = B[i - 1] * i % mo;
  }
  a = A[2 * n + 1][0][1] * B[n] % mo * f(2, n) % mo;
  b = f(B[2 * n + 1], mo - 2) * l % mo;
  cout << a * b % mo << endl;
}