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;
using LL = long long;
constexpr const int kMod = 998244353;
LL coeff[4000];
LL nCr[2003][2003];
void ComputeComb() {
  nCr[0][0] = 1;
  for (int i = 1; i <= 2000; i++) {
    nCr[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      nCr[i][j] = (nCr[i - 1][j - 1] + nCr[i - 1][j]) % kMod;
    }
  }
}
LL modexp(LL a, LL b) {
  LL res = 1;
  while (b) {
    if ((b & 1)) {
      res *= a;
      res %= kMod;
    }
    a *= a;
    a %= kMod;
    b >>= 1;
  }
  return res;
}
int main() {
  ComputeComb();
  int N, K;
  LL L;
  cin >> N >> K >> L;
  for (int k = 0; k <= N - K; k++) {
    for (int i = 0; i <= k; i++) {
      LL cs = 1;
      if ((k - i) % 2 != 0) {
        cs = kMod - 1;
      }
      cs *= nCr[k][i];
      cs %= kMod;
      cs *= nCr[N][k];
      cs %= kMod;
      coeff[N - i] += cs;
      coeff[N - i] %= kMod;
    }
  }
  LL ans = 0;
  for (int g = 0; g <= N; g++) {
    LL csum = 0;
    for (int i = 0; i <= g; i++) {
      LL cs = 1;
      if ((g - i) % 2 == 1) {
        cs = kMod - 1;
      }
      cs *= nCr[g][i];
      cs %= kMod;
      csum += (cs * modexp(2 * g - i + 1, kMod - 2)) % kMod;
    }
    csum %= kMod;
    csum *= modexp(2, g);
    csum %= kMod;
    csum *= coeff[g];
    csum %= kMod;
    ans += csum;
    ans %= kMod;
  }
  ans *= L;
  ans %= kMod;
  cout << ans << endl;
  return 0;
}