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>
const int INF = 0x3f3f3f3f;
const int MOD = 998244353;
const int ny2 = (MOD + 1) / 2;
const int N = 4005;
const long double pi = acos(-1);
const long double e = exp(1);
const long double eps = 1e-8;
long long fac[N], f[N][N];
int read() {
  int x = 0, v = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; v = (ch == '-') ? (-1) : v, ch = getchar())
    ;
  for (; ch <= '9' && ch >= '0'; x = x * 10 + ch - '0', ch = getchar())
    ;
  return x * v;
}
void upd(long long &x, long long v) { x += v, (x >= MOD) ? (x -= MOD) : 0; }
long long ksm(long long x, long long dep) {
  long long res = 1;
  for (; dep; dep >>= 1, x = x * x % MOD) {
    (dep & 1) ? (res = res * x % MOD) : 0;
  }
  return res;
}
int main(void) {
  fac[0] = fac[1] = 1;
  for (int i = 2; i <= N - 1; ++i) fac[i] = fac[i - 1] * i % MOD;
  int n = read(), k = read(), l = read();
  f[1][1] = 1;
  for (int i = 1; i <= n * 2; ++i)
    for (int j = 0; j <= std::min(i, n); ++j) {
      if (!f[i][j]) continue;
      upd(f[i + 1][j + 1], f[i][j]);
      if (j) upd(f[i + 1][j - 1], f[i][j] * j % MOD);
    }
  long long ans = 0;
  for (int i = 1; i <= n * 2; ++i)
    for (int j = k; j <= std::min(i, n); ++j) {
      upd(ans, f[i][j] * f[n * 2 - i][j] % MOD * fac[j] % MOD);
    }
  ans = ans * ksm(f[n * 2][0], MOD - 2) % MOD;
  ans = ans * l % MOD * ksm(n * 2 + 1, MOD - 2) % MOD;
  printf("%lld\n", ans);
  return 0;
}