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 N = 5011;
const int mod = 998244353;
int n, k, l;
long long C[N][N];
long long a[N], b[N], help[N], v[N];
long long calc(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y & 1) (z *= x) %= mod;
    (x *= x) %= mod, y /= 2;
  }
  return z;
}
long long ni(long long x) { return calc(x, mod - 2); }
int main() {
  scanf("%d %d %d", &n, &k, &l);
  C[0][0] = 1;
  for (int i = 1; i < N; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  }
  for (int i = k; i <= n; i++) {
    long long V = C[n][i] % mod * calc(mod - 1, i) % mod;
    long long l2_2 = ni(2) * l % mod * l % mod;
    for (int j = 0; j <= n - i; j++) {
      v[j + i] += C[n - i][j] * calc(l2_2, n - i - j) % mod * V % mod;
      if (v[j + i] >= mod) v[j + i] -= mod;
    }
  }
  b[0] = 1;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n * 2; j++) (a[j] += v[i] * b[j]) %= mod;
    for (int j = 0; j <= n * 2; j++) {
      (help[j + 1] += (mod - l) * b[j] % mod) %= mod;
      (help[j + 2] += b[j]) %= mod;
    }
    for (int j = 0; j <= n * 2 + 10; j++) b[j] = help[j], help[j] = 0;
  }
  for (int i = 0; i <= n * 2; i++) help[i + 1] = a[i] * ni(i + 1) % mod;
  long long ans = 0;
  for (int i = 0; i <= 2 * n + 1; i++) (ans += help[i] * calc(l, i)) %= mod;
  printf("%d\n", ans * ni(calc(l, 2 * n)) % mod * calc(2, n) % mod);
  return 0;
}