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 readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int cys = 998244353;
long long n, k, l;
long long fac[10005], ifac[10005], pw[10005];
long long qpow(long long x, long long p) {
  long long ret = 1;
  for (; p; p >>= 1, x = x * x % cys)
    if (p & 1) ret = ret * x % cys;
  return ret;
}
int main() {
  n = readint();
  k = readint();
  l = readint();
  fac[0] = ifac[0] = pw[0] = 1;
  for (int i = 1; i <= 2 * n + 1; i++)
    fac[i] = fac[i - 1] * i % cys, pw[i] = pw[i - 1] * 2 % cys;
  ifac[2 * n + 1] = qpow(fac[2 * n + 1], cys - 2);
  for (int i = 2 * n; i >= 1; i--) ifac[i] = ifac[i + 1] * (i + 1) % cys;
  long long ans = l * fac[n] % cys, sum = 0;
  for (int i = k; i <= n; i++)
    for (int j = 0; j <= n - i; j++)
      sum = (sum + ifac[i] * ifac[j] % cys * pw[i + j] % cys * fac[i + j] %
                       cys * fac[i + j] % cys * ifac[2 * i + 2 * j + 1] % cys *
                       ifac[n - i - j] % cys * ((j & 1) ? cys - 1 : 1) % cys) %
            cys;
  printf("%lld\n", ans * sum % cys);
  return 0;
}