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 = 4003, mod = 998244353;
int fac[N], inv[N], invfac[N], po[N];
inline void init(int m) {
  fac[0] = 1;
  for (register int i = 1; i <= m; i++)
    fac[i] = (long long)i * fac[i - 1] % mod;
  inv[1] = 1;
  for (register int i = 2; i <= m; i++)
    inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod;
  invfac[0] = 1;
  for (register int i = 1; i <= m; i++)
    invfac[i] = (long long)invfac[i - 1] * inv[i] % mod;
  po[0] = 1;
  for (register int i = 1; i <= m; i++) po[i] = (po[i - 1] << 1) % mod;
}
inline int C(int n, int m) {
  if (n < m || m < 0) return 0;
  return (long long)fac[n] * invfac[n - m] % mod * invfac[m] % mod;
}
int n, k, l, ans, f[N];
int main() {
  scanf("%d%d%d", &n, &k, &l);
  init(n << 1 | 1);
  for (register int i = k; i <= n; i++)
    for (register int j = 0; j <= i; j++) {
      int t = (long long)C(i, j) * inv[i + j + 1] % mod;
      if (j & 1)
        f[i] = (f[i] + mod - t) % mod;
      else
        f[i] = (f[i] + t) % mod;
    }
  for (register int i = k; i <= n; i++) {
    int t1 = 0;
    for (register int j = 0; j <= n - i; j++) {
      int t2 = (long long)po[i + j] * C(n - i, j) % mod * f[i + j] % mod;
      if (j & 1)
        t1 = (t1 + mod - t2) % mod;
      else
        t1 = (t1 + t2) % mod;
    }
    ans = (ans + (long long)t1 * C(n, i) % mod) % mod;
  }
  printf("%d", (long long)l * ans % mod);
}