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 mod = 998244353;
int n, k, l;
int f[10200];
int fac[10200], inv[10200], pw[10200];
int C(int nn, int kk) {
  return 1ll * fac[nn] * inv[kk] % mod * inv[nn - kk] % mod;
}
int A(int nn, int kk) { return 1ll * fac[nn] * inv[nn - kk] % mod; }
int fast_pow(int now, int pw) {
  int ans = 1, dt = now, bit = 1;
  while (bit <= pw) {
    if (bit & pw) {
      ans = 1ll * ans * dt % mod;
    }
    dt = 1ll * dt * dt % mod;
    bit <<= 1;
  }
  return ans;
}
int work(int rm) {
  int ll = 0, r = 2 * n;
  int Rp = fast_pow(f[2 * n], mod - 2);
  int ans = 0;
  while (r >= 0) {
    int now = C(ll, rm), im = A(r, rm);
    int res = 1ll * now * im % mod * f[ll - rm] % mod * f[r - rm] % mod *
              pw[rm] % mod;
    res = 1ll * res * Rp % mod;
    ans += res;
    ans %= mod;
    ll++;
    r--;
  }
  ans = 1ll * ans * fast_pow(2 * n + 1, mod - 2) % mod * l % mod;
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &k, &l);
  f[0] = 1;
  fac[0] = 1;
  pw[0] = 1;
  for (int i = 1; i <= 2 * n; i++)
    pw[i] = 2 * pw[i - 1] % mod, fac[i] = 1ll * fac[i - 1] * i % mod;
  for (int i = 2; i <= 2 * n; i++) {
    f[i] = 2ll * f[i - 2] % mod * (i - 1) % mod;
  }
  inv[2 * n] = fast_pow(fac[2 * n], mod - 2);
  for (int i = 2 * n - 1; i >= 0; i--)
    inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  int ans = 0;
  for (int i = k; i <= n; i++) ans += work(i), ans %= mod;
  printf("%d\n", ans);
  return 0;
}