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;
inline void open(const char *s) {}
inline int rd() {
  static int x, f;
  x = 0;
  f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return f > 0 ? x : -x;
}
const int N = 2010, mod = 998244353;
int n, m, L, ans, f[N << 1][N], fac[N << 1];
inline int pls(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int mul(int a, int b) { return (long long)a * b % mod; }
inline int pwr(int x, int k) {
  int res = 1;
  for (; k; k >>= 1, x = mul(x, x))
    if (k & 1) res = mul(res, x);
  return res;
}
int main() {
  n = rd();
  m = rd();
  L = rd();
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mul(fac[i - 1], i);
  f[1][1] = 1;
  for (int i = 1; i <= (n << 1) - 1; i++) {
    for (int j = 0; j <= min(i, n); j++) {
      if (!f[i][j]) continue;
      if (j + 1 <= n * 2 - i - 1)
        f[i + 1][j + 1] = pls(f[i + 1][j + 1], f[i][j]);
      if (j) f[i + 1][j - 1] = pls(f[i + 1][j - 1], mul(j, f[i][j]));
    }
  }
  int ans = 0;
  for (int i = 1; i <= (n << 1) - 1; i++)
    for (int j = m; j <= n; j++) {
      ans = pls(ans, mul(f[i][j], mul(f[n * 2 - i][j], fac[j])));
    }
  ans = mul(ans, pwr(f[n << 1][0], mod - 2));
  ans = mul(ans, L);
  ans = mul(ans, pwr(n * 2 + 1, mod - 2));
  cout << ans;
  return 0;
}