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 = 100000;
const double pi = acos(-1.0);
int n, k, l;
long long dp[4020][4020], fac[4040];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while ((ch < '0') || (ch > '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ((ch >= '0') && (ch <= '9')) {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
long long qpow(long long x, int y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = (ans * x) % 998244353;
    x = (x * x) % 998244353;
    y >>= 1;
  }
  return ans;
}
long long inv(long long x) { return qpow(x, 998244353 - 2); }
int main() {
  n = read();
  k = read();
  l = read();
  dp[0][0] = 1;
  for (int i = 0; i <= n * 2; i++) {
    for (int j = 0; j <= min(i, n); j++) {
      dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j]) % 998244353;
      dp[i + 1][j - 1] = (dp[i + 1][j - 1] + dp[i][j] * j) % 998244353;
    }
  }
  fac[0] = 1;
  for (int i = 1; i <= 4000; i++) fac[i] = (fac[i - 1] * i) % 998244353;
  long long ans = 0;
  for (int i = 1; i <= n * 2; i++) {
    for (int j = k; j <= min(i, n); j++) {
      ans = (ans +
             ((dp[i][j] * dp[n * 2 - i][j]) % 998244353 * fac[j]) % 998244353) %
            998244353;
    }
  }
  long long len = l * inv(n * 2 + 1) % 998244353;
  ans = ans * inv(dp[n * 2][0]) % 998244353;
  ans = ans * len % 998244353;
  printf("%lld", ans);
  return 0;
}