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 P = 998244353;
const int MAXN = 4050;
int n, K, l;
inline int add(int x, int y) {
  x += y;
  if (x >= P) x -= P;
  return x;
}
inline long long ksm(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % P;
    x = x * x % P;
    y >>= 1;
  }
  return ans;
}
long long fac[MAXN], ifac[MAXN];
inline void init() {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= 2 * n + 1; ++i) fac[i] = fac[i - 1] * i % P;
  ifac[2 * n + 1] = ksm(fac[2 * n + 1], P - 2);
  for (int i = 2 * n; i >= 1; --i) ifac[i] = ifac[i + 1] * (i + 1) % P;
}
int f[MAXN][MAXN][2];
int main() {
  scanf("%d%d%d", &n, &K, &l);
  init();
  f[0][0][0] = 1;
  for (int i = 1; i <= 2 * n + 1; ++i) {
    for (int j = 0; j <= i; ++j) {
      for (int k = 0; k <= 1; ++k) {
        f[i][j][k] = add(f[i][j][k], 1ll * (j + 1) * f[i - 1][j + 1][k] % P);
        f[i][j][k] = add(f[i][j][k], f[i - 1][j - 1][k]);
        if (k && j >= K) f[i][j][k] = add(f[i][j][k], f[i - 1][j][k ^ 1]);
      }
    }
  }
  printf("%lld\n", 1ll * f[2 * n + 1][0][1] * ksm(2, n) % P * fac[n] % P *
                       ifac[2 * n + 1] % P * l % P);
  return 0;
}