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 inf = 0x3f3f3f3f;
const int Mod = 998244353;
const int M = 4005;
const int N = 2005;
int read() {
  int s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s * f;
}
int n, m, L, f[M][N][2], fac[M], inv[M];
int mod(int x) { return x >= Mod ? x - Mod : x; }
int Pow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % Mod;
    x = 1ll * x * x % Mod;
    y >>= 1;
  }
  return ans;
}
void Prepare() {
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i < M; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
  for (int i = 2; i < M; i++)
    inv[i] = 1ll * inv[Mod % i] * (Mod - Mod / i) % Mod;
  for (int i = 2; i < M; i++) inv[i] = 1ll * inv[i - 1] * inv[i] % Mod;
}
int main() {
  n = read();
  m = read();
  L = read();
  Prepare();
  f[0][0][0] = 1;
  for (int i = 0; i < n * 2 + 1; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k < 2; k++) {
        f[i + 1][j + 1][k] = mod(f[i + 1][j + 1][k] + f[i][j][k]);
        if (j)
          f[i + 1][j - 1][k] =
              mod(f[i + 1][j - 1][k] + 1ll * f[i][j][k] * j % Mod);
      }
      if (j >= m) f[i + 1][j][1] = mod(f[i + 1][j][1] + f[i][j][0]);
    }
  }
  int Ans = 1ll * f[n * 2 + 1][0][1] * Pow(2, n) % Mod * fac[n] % Mod *
            inv[n * 2 + 1] % Mod * L % Mod;
  printf("%d\n", Ans);
  return 0;
}