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;
template <typename T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
const int INF = 0x3f3f3f3f;
const int N = 2e3 + 10, mod = 998244353;
int dp[N << 1][N << 1], fac[N << 1];
inline int read() {
  int x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') flag = -1, ch = getchar();
  while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch - '0'), ch = getchar();
  return x * flag;
}
inline int fpm(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod, b /= 2;
  }
  return res;
}
int main() {
  int n = read(), k = read(), l = read(), ans = 0;
  fac[0] = 1;
  for (int i = (1), iend = (n); i <= iend; i++)
    fac[i] = 1ll * fac[i - 1] * i % mod;
  dp[0][0] = 1;
  for (int i = (1), iend = (n << 1); i <= iend; i++)
    for (int j = (0), jend = (i); j <= jend; j++) {
      if (j) dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % mod;
      dp[i][j] = (dp[i][j] + 1ll * dp[i - 1][j + 1] * (j + 1)) % mod;
    }
  for (int i = (0), iend = ((n << 1)); i <= iend; i++)
    for (int j = (k), jend = (i); j <= jend; j++)
      ans = (ans + 1ll * dp[i][j] * dp[(n << 1) - i][j] % mod * fac[j]) % mod;
  printf("%lld\n", 1ll * ans * l % mod * fpm(2 * n + 1, mod - 2) % mod *
                       fpm(dp[n << 1][0], mod - 2) % mod);
  return 0;
}