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 MOD = 998244353;
long long modpow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      ans *= a;
      ans %= MOD;
    }
    a *= a;
    a %= MOD;
    b /= 2;
  }
  return ans;
}
long long modinv(long long a) { return modpow(a, MOD - 2); }
int main() {
  int n, k, l;
  cin >> n >> k >> l;
  vector<vector<long long>> dp1(n * 2 + 1, vector<long long>(n + 1, 0));
  vector<vector<long long>> dp2(n * 2 + 1, vector<long long>(n + 1, 0));
  dp1[0][0] = 1;
  for (int i = 0; i < n * 2; i++) {
    for (int j = 0; j <= n; j++) {
      dp1[i + 1][j] += dp1[i][j] * (2 * j - i);
      dp1[i + 1][j] %= MOD;
      dp2[i + 1][j] += dp2[i][j] * (2 * j - i);
      if (2 * j - (i + 1) >= k) {
        dp2[i + 1][j] += dp1[i][j] * (2 * j - i);
      }
      dp2[i + 1][j] %= MOD;
      if (j < n) {
        dp1[i + 1][j + 1] += dp1[i][j];
        dp1[i + 1][j + 1] %= MOD;
        dp2[i + 1][j + 1] += dp2[i][j];
        if (2 * (j + 1) - (i + 1) >= k) {
          dp2[i + 1][j + 1] += dp1[i][j];
        }
        dp2[i + 1][j + 1] %= MOD;
      }
    }
  }
  long long L = l * modinv(n * 2 + 1) % MOD;
  cout << dp2[n * 2][n] * L % MOD * modinv(dp1[n * 2][n]) % MOD << endl;
}