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 md = 998244353, N = 20005;
int n, K, l, i, j, k, f[N], g[N], h[N], ans, inv[N], jc[N], rjc[N];
int C(int n, int m) { return 1ll * jc[n] * rjc[m] % md * rjc[n - m] % md; }
int main() {
  scanf("%d%d%d", &n, &K, &l);
  inv[1] = 1;
  for (i = 2; i <= 20000; ++i) inv[i] = 1ll * inv[md % i] * (md - md / i) % md;
  jc[0] = rjc[0] = 1;
  for (i = 1; i <= 20000; ++i)
    jc[i] = 1ll * jc[i - 1] * i % md, rjc[i] = 1ll * rjc[i - 1] * inv[i] % md;
  f[0] = 1;
  for (i = 1; i <= n; ++i) {
    for (j = 2 * i; j >= 0; --j) {
      if (j >= 1) f[j] = (f[j] - f[j - 1] * 2ll % md + md) % md;
      if (j >= 2) f[j] = (f[j] + f[j - 2] * 2ll % md) % md;
    }
  }
  for (i = 0;; ++i) {
    if (i >= K) {
      k = C(n, i);
      for (j = 0; j <= 2 * n; ++j) g[j] = (g[j] + 1ll * k * f[j]) % md;
    }
    if (i == n) break;
    for (j = 1; j <= 2 * n; ++j) {
      f[j] = (f[j] + f[j - 1] * 2ll) % md;
      if (j >= 2) f[j] = (f[j] - f[j - 2] * 2ll % md + md) % md;
    }
    for (j = 2 * n; j >= 0; --j) {
      f[j] = 0;
      if (j >= 1) f[j] = (f[j] + 2ll * f[j - 1]) % md;
      if (j >= 2) f[j] = (f[j] - 2ll * f[j - 2] % md + md) % md;
    }
  }
  for (j = 0; j <= 2 * n; ++j) ans = (ans + 1ll * g[j] * inv[j + 1]) % md;
  ans = 1ll * ans * l % md;
  printf("%d", ans);
  return 0;
}