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 maxn = 2e3 + 100;
const long long mod = 998244353;
const long long inf = 1e18;
long long n, k, l, dp[2 * maxn][maxn], s[2 * maxn][maxn];
long long pw(long long pa, long long push_back) {
  if (push_back == 0) return 1;
  long long pc = pw(pa, push_back / 2);
  pc = (pc * pc) % mod;
  if (push_back % 2) pc = (pc * pa) % mod;
  return pc;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k >> l;
  dp[0][0] = 1;
  for (long long i = 1; i <= 2 * n; i++) {
    for (long long j = 0; j + 1 < maxn; j++) {
      if (j) dp[i][j] = dp[i - 1][j - 1], s[i][j] = s[i - 1][j - 1];
      dp[i][j] = (dp[i][j] + dp[i - 1][j + 1] * (j + 1)) % mod;
      s[i][j] = (s[i][j] + s[i - 1][j + 1] * (j + 1)) % mod;
      if (j >= k) s[i][j] = (s[i][j] + dp[i][j]) % mod;
    }
  }
  cout << (((((s[2 * n][0] * pw(dp[2 * n][0], mod - 2)) % mod) * l) % mod) *
           pw(2 * n + 1, mod - 2)) %
              mod;
  return 0;
}