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 b, long long e, long long m) {
  long long r = 1;
  while (e) {
    if (e % 2) {
      r *= b;
      r %= m;
    }
    b *= b;
    b %= m;
    e /= 2;
  }
  return r;
}
long long inv(long long b) { return modpow(b, MOD - 2, MOD); }
void solve() {
  int n, k, l;
  cin >> n >> k >> l;
  vector<long long> dp;
  long long divdiv = inv(2 * n + 1);
  dp.resize(n + 1);
  dp[0] = 1;
  long long ret = 0;
  for (int e = 2 * n; e > 0; e--) {
    vector<long long> ndp;
    ndp.resize(n + 1);
    long long denom = inv(e);
    for (int i = 0; i < dp.size(); i++) {
      if (dp[i] == 0) continue;
      if (i >= k) {
        ret += dp[i];
        ret %= MOD;
      }
      int canOpen = e - i;
      int canClose = i;
      if (i + 1 < ndp.size()) {
        long long inc = canOpen;
        inc *= denom;
        inc %= MOD;
        inc *= dp[i];
        inc %= MOD;
        ndp[i + 1] += inc;
        ndp[i + 1] %= MOD;
      }
      if (i - 1 >= 0) {
        long long inc = canClose;
        inc *= denom;
        inc %= MOD;
        inc *= dp[i];
        inc %= MOD;
        ndp[i - 1] += inc;
        ndp[i - 1] %= MOD;
      }
    }
    dp = ndp;
  }
  ret *= l;
  ret %= MOD;
  ret *= divdiv;
  ret %= MOD;
  cout << ret << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}