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;
inline int read() {
  int x = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * w;
}
const int mod = 998244353;
inline int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % mod;
    x = 1ll * x * x % mod;
    y = y >> 1;
  }
  return ret;
}
inline int inv(int x) { return qpow(x, mod - 2); }
int n, k, l;
int f[4005][4005], jc[4005], ans;
int main() {
  cin >> n >> k >> l;
  l = 1ll * l * inv(n * 2 + 1) % mod;
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = 1ll * jc[i - 1] * i % mod;
  f[0][0] = 1;
  for (int i = 0; i < n * 2; i++)
    for (int j = min(n, i); j >= 0; j--) {
      f[i + 1][j + 1] = (f[i + 1][j + 1] + f[i][j]) % mod;
      if (j) f[i + 1][j - 1] = (f[i + 1][j - 1] + 1ll * f[i][j] * j) % mod;
    }
  for (int i = 1; i <= n * 2; i++)
    for (int j = k; j <= n; j++)
      ans = (ans + 1ll * f[i][j] * f[n * 2 - i][j] % mod * jc[j]) % mod;
  cout << 1ll * ans * l % mod * inv(f[n * 2][0]) % mod;
  return 0;
}