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;
int _max(int x, int y) { return x > y ? x : y; }
int _min(int x, int y) { return x < y ? x : y; }
const int mod = 998244353;
const int N = 2001;
int read() {
  int s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * f;
}
void put(int x) {
  if (x >= 10) put(x / 10);
  putchar(x % 10 + '0');
}
int f[N * 2][N + 1], jc[N];
int add(int x, int y) {
  x += y;
  return x >= mod ? x - mod : x;
}
int pow_mod(int a, int k) {
  int ans = 1;
  while (k) {
    if (k & 1) ans = (long long)ans * a % mod;
    a = (long long)a * a % mod, k /= 2;
  }
  return ans;
}
int main() {
  int n = read(), k = read(), L = read();
  f[0][0] = 1;
  for (int i = 1; i <= 2 * n; i++)
    for (int j = 0; j <= _min(i, n); j++) {
      f[i][j] = add((long long)f[i - 1][j + 1] * (j + 1) % mod,
                    j == 0 ? 0 : f[i - 1][j - 1]);
    }
  int ans = 0;
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = (long long)jc[i - 1] * i % mod;
  for (int i = 1; i <= 2 * n; i++)
    for (int j = k; j <= _min(2 * n - i, i); j++) {
      ans = add(ans, (long long)f[i][j] * f[2 * n - i][j] % mod * jc[j] % mod);
    }
  put((long long)ans * L % mod * pow_mod(2 * n + 1, mod - 2) % mod *
      pow_mod(f[2 * n][0], mod - 2) % mod);
  return 0;
}