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 mod = 998244353, n, m, f, ans;
long long dp[4005][4005][2], fakt[4005];
long long oszt(long long a) {
  long long ans = 1, p = mod - 2;
  while (p) {
    if (p % 2) ans *= a, ans %= mod;
    a *= a, a %= mod, p /= 2;
  }
  return ans;
}
int main() {
  cin >> n >> m >> ans, fakt[0] = 1, f = 2 * n + 1;
  for (long long i = 1; i <= f; i++) fakt[i] = (fakt[i - 1] * i) % mod;
  dp[0][0][0] = 1;
  for (long long i = 0; i <= f; i++) {
    for (long long j = 0; j <= min(i, n); j++) {
      long long x = dp[i][j][0] % mod, y = dp[i][j][1] % mod;
      dp[i + 1][j + 1][0] += x, dp[i + 1][j + 1][1] += y;
      if (j) dp[i + 1][j - 1][0] += x * j, dp[i + 1][j - 1][1] += y * j;
      if (j >= m) dp[i + 1][j][1] += x;
    }
  }
  for (long long i = 1; i <= n; i++) ans *= 2, ans %= mod;
  ans *= fakt[n], ans %= mod, ans *= oszt(fakt[f]), ans %= mod,
      ans *= dp[f][0][1], ans %= mod;
  cout << ans << "\n";
  return 0;
}