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 mu = 998244353;
int dp[4010][4010][2], n, k, l, fac[5010];
void reduce(int &x) { x += x >> 31 & mu; }
int ksm(int x, int y, int ans = 1) {
  for (; y; y >>= 1, x = 1ll * x * x % mu)
    if (y & 1) ans = 1ll * ans * x % mu;
  return ans;
}
int main() {
  scanf("%d %d %d", &n, &k, &l);
  dp[0][0][0] = 1;
  for (int i = 0; i <= n * 2; i++)
    for (int j = 0; j <= i; j++)
      for (int o = 0; o <= 1; o++) {
        reduce(dp[i + 1][j + 1][o] += dp[i][j][o] - mu);
        if (o == 0 && j >= k) reduce(dp[i + 1][j][1] += dp[i][j][o] - mu);
        if (j) reduce(dp[i + 1][j - 1][o] += 1ll * dp[i][j][o] * j % mu - mu);
      }
  fac[0] = 1;
  for (int i = 1; i <= n * 2 + 1; i++) fac[i] = 1ll * fac[i - 1] * i % mu;
  printf("%d\n", 1ll * dp[n * 2 + 1][0][1] * ksm(2, n) % mu * fac[n] % mu *
                     ksm(fac[n * 2 + 1], mu - 2) % mu * l % mu);
  return 0;
}