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;
long long fpow(long long a, long long b) {
  long long ans = 1, t = a;
  while (b) {
    if (b & 1) ans = ans * t % 998244353;
    t = t * t % 998244353;
    b >>= 1;
  }
  return ans;
}
long long inv(long long a) { return fpow(a, 998244353 - 2); }
inline void add(int& a, int b) {
  a += b;
  if (a >= 998244353) a -= 998244353;
}
inline int P(int x) { return x & 1; }
int dp[2][4010][2];
int fac[5010];
int main() {
  int n, k, l;
  scanf("%d %d %d", &n, &k, &l);
  dp[0][0][0] = 1;
  for (int i = 0; i <= n * 2; i++) {
    memset(dp[P(i + 1)], 0, sizeof(dp[P(i + 1)]));
    for (int j = 0; j <= i; j++)
      for (int o = 0; o <= 1; o++) {
        add(dp[P(i + 1)][j + 1][o], dp[P(i)][j][o]);
        if (o == 0 && j >= k) add(dp[P(i + 1)][j][1], dp[P(i)][j][o]);
        if (j)
          add(dp[P(i + 1)][j - 1][o],
              (long long)dp[P(i)][j][o] * j % 998244353);
      }
  }
  fac[0] = 1;
  for (int i = 1; i <= n * 2 + 1; i++)
    fac[i] = (long long)fac[i - 1] * i % 998244353;
  printf("%lld\n", (long long)dp[P(n * 2 + 1)][0][1] * fpow(2, n) % 998244353 *
                       fac[n] % 998244353 * inv(fac[n * 2 + 1]) % 998244353 *
                       l % 998244353);
  return 0;
}