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 = 4010, mod = 998244353;
long long dp[N][N], jc[N] = {1};
void Add(long long &x, long long y) { x = (x + y) % mod; }
void Mul(long long &x, long long y) { x = (x * y) % mod; }
long long add(long long x, long long y) { return (x + y) % mod; }
long long mul(long long x, long long y) { return x * y % mod; }
long long q_pow(long long x, int y = mod - 2) {
  long long ans = 1;
  for (; y; y >>= 1) {
    if (y & 1) Mul(ans, x);
    Mul(x, x);
  }
  return ans;
}
int main() {
  int n, k, l;
  scanf("%d%d%d", &n, &k, &l);
  dp[1][1] = 1;
  for (int i = 1; i <= n * 2; ++i)
    for (int j = 0; j <= min(i, n); ++j) {
      if (dp[i][j]) {
        Add(dp[i + 1][j + 1], dp[i][j]);
        if (j) Add(dp[i + 1][j - 1], mul(dp[i][j], j));
      }
    }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) jc[i] = mul(jc[i - 1], i);
  for (int i = 1; i <= n * 2; ++i)
    for (int j = k; j <= min(i, n); ++j) {
      Add(ans, mul(mul(dp[i][j], dp[n * 2 - i][j]), jc[j]));
    }
  Mul(ans, mul(l, q_pow(2 * n + 1)));
  Mul(ans, q_pow(dp[2 * n][0]));
  printf("%lld\n", ans);
  return 0;
}