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;
struct modint {
  long long const static mod = 998244353;
  long long a = 0;
  modint() {}
  modint(long long x) { *this = x; }
  void operator=(long long x) {
    a = x % mod;
    if (a < 0) a += mod;
  }
  bool operator==(modint y) { return a == y.a; }
  bool operator!=(modint y) { return a != y.a; }
  modint operator+(modint y) { return modint(a + y.a); }
  modint operator-(modint y) { return modint(a - y.a + mod); }
  modint operator*(modint y) { return modint(a * y.a); }
  modint pow(modint x, int y) {
    if (!y) return modint(1);
    if (y % 2) return x * pow(x * x, y / 2);
    return pow(x * x, y / 2);
  }
  modint operator/(modint y) { return *this * pow(y, mod - 2); }
  modint operator^(long long x) {
    if (x == 0) return 1;
    return (x % 2 ? *this : modint(1)) * (((*this) * (*this)) ^ (x / 2));
  }
};
int n, k, l;
modint dp[2005][2005];
modint inv[10000];
int main() {
  cin >> n >> k >> l;
  for (int i = 1; i <= 10000; i++) inv[i] = modint(1) / modint(i);
  dp[0][0] = 1;
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < n + 1; j++) {
      if (i > 0)
        dp[i][j] = dp[i][j] + dp[i - 1][j] * modint(2 * (n - i + 1)) *
                                  inv[2 * (n - i + 1) + (i - 1 - j)];
      if (j > 0)
        dp[i][j] = dp[i][j] + dp[i][j - 1] * modint(i - j + 1) *
                                  inv[2 * (n - i) + (i - j + 1)];
    }
  }
  modint ans = 0;
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < n + 1; j++) {
      if (i - j >= k) ans = ans + dp[i][j];
    }
  }
  ans = ans * modint(l) / modint(2 * n + 1);
  cout << ans.a << endl;
}