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 MAXN = 2005;
const int Mod = 998244353;
int fp(int a, int k) {
  if (k < 0) a = fp(a, Mod - 2), k = -k;
  int res = 1;
  while (k) {
    if (k & 1) res = 1LL * res * a % Mod;
    a = 1LL * a * a % Mod;
    k >>= 1;
  }
  return res;
}
int C[MAXN][MAXN], Inv[MAXN << 1];
int D[MAXN][MAXN << 1], E[MAXN];
int main() {
  int n, k, l;
  scanf("%d%d%d", &n, &k, &l);
  for (int i = (C[0][0] = 1); i <= n; i++)
    for (int j = (C[i][0] = 1); j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % Mod;
  D[0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= 2 * i; j++) {
      D[i + 1][j + 2] = (D[i + 1][j + 2] - 2LL * D[i][j] % Mod + Mod) % Mod;
      D[i + 1][j + 1] = (D[i + 1][j + 1] + 2LL * D[i][j]) % Mod;
    }
  for (int i = 1; i <= 2 * n + 1; i++) Inv[i] = fp(i, -1);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= 2 * i; j++)
      E[i] = (E[i] + 1LL * D[i][j] * Inv[j + 1]) % Mod;
  int res = 0;
  for (int t = k; t <= n; t++) {
    int tmp = 0;
    for (int i = 0; i <= n - t; i++) {
      int mul = 1LL * C[n - t][i] * E[i + t] % Mod;
      if (~i & 1)
        tmp = (tmp + mul) % Mod;
      else
        tmp = (tmp - mul + Mod) % Mod;
    }
    res = (res + 1LL * C[n][t] * tmp) % Mod;
  }
  res = 1LL * res * l % Mod;
  printf("%d\n", res);
  return 0;
}