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>
const int z = 998244353;
int fp(int a, int x) {
  if (x == 0) {
    return 1;
  }
  if (x % 2 == 1) {
    return 1ll * a * fp(a, x - 1) % z;
  }
  int t = fp(a, x / 2);
  return 1ll * t * t % z;
}
int inv(int a) { return fp(a, z - 2); }
int main() {
  int n, k, l;
  scanf("%d %d %d", &n, &k, &l);
  int pow2[2 * n + 1], ifac[2 * n + 1];
  pow2[0] = ifac[0] = 1;
  for (int i = 1; i <= 2 * n; i++) {
    pow2[i] = 2ll * pow2[i - 1] % z;
    ifac[i] = 1ll * ifac[i - 1] * inv(i) % z;
  }
  int facn = inv(ifac[n]);
  int fac2n = inv(ifac[2 * n]);
  int ans = 0;
  for (int i = 0; i <= 2 * n; i++) {
    int invi = inv(1ll * fac2n * ifac[i] % z * ifac[2 * n - i] % z);
    for (int j = k; j <= n; j++) {
      int p;
      if ((i + j) % 2 != 0 || j > i || i + j > 2 * n) {
        p = 0;
      } else {
        int k = (i + j) / 2;
        p = 1ll * facn * ifac[n - k] % z * ifac[i - k] % z * ifac[2 * k - i] %
            z;
        p = 1ll * p * invi % z * pow2[2 * k - i] % z;
      }
      ans = (ans + p) % z;
    }
  }
  ans = 1ll * ans * l % z * inv(2 * n + 1) % z;
  printf("%d\n", ans);
  return 0;
}