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 long long M = 998244353;
long long dp[4051][4051][2];
long long factorial(int l, int r) {
  long long ans = 1;
  for (int i = l; i <= r; ++i) ans = (ans * i) % M;
  return ans;
}
long long mi(long long x, long long y) {
  long long base = x;
  long long ans = 1;
  while (y != 0) {
    if (y & 1) ans = (ans * base) % M;
    base = (base * base) % M;
    y >>= 1;
  }
  return ans;
}
int main() {
  int n, k, l;
  scanf("%d%d%d", &n, &k, &l);
  l %= M;
  dp[0][0][0] = 1;
  for (int i = 1; i <= 2 * n + 1; ++i) {
    for (int j = 0; j <= i; ++j) {
      for (int m = 0; m <= 1; ++m) {
        if (j - 1 >= 0) dp[i][j][m] = (dp[i][j][m] + dp[i - 1][j - 1][m]) % M;
        dp[i][j][m] = (dp[i][j][m] + dp[i - 1][j + 1][m] * (j + 1) % M) % M;
      }
      if (j >= k) dp[i][j][1] = (dp[i][j][1] + dp[i - 1][j][0]) % M;
    }
  }
  long long ans = mi(factorial(n + 1, 2 * n + 1), M - 2) * dp[2 * n + 1][0][1] %
                  M * mi(2, n) % M * l % M;
  printf("%lld", ans);
}