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 int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-12;
const double PI = acos(-1.0);
const double angcst = PI / 180.0;
const long long mod = 998244353;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long qmul(long long a, long long b) {
  long long r = 0;
  while (b) {
    if (b & 1) r = (r + a) % mod;
    b >>= 1;
    a = (a + a) % mod;
  }
  return r;
}
long long qpow(long long a, long long n) {
  long long r = 1;
  while (n) {
    if (n & 1) r = (r * a) % mod;
    n >>= 1;
    a = (a * a) % mod;
  }
  return r;
}
long long qpow(long long a, long long n, long long p) {
  long long r = 1;
  while (n) {
    if (n & 1) r = (r * a) % p;
    n >>= 1;
    a = (a * a) % p;
  }
  return r;
}
long long fac[5050], inv[5050], n2[5050];
long long tinv[5050];
void init() {
  fac[0] = inv[0] = n2[0] = 1;
  for (int i = (1); i <= (5000); i++)
    fac[i] = fac[i - 1] * i % mod, n2[i] = (n2[i - 1] << 1) % mod,
    tinv[i] = qpow(i, mod - 2);
  inv[5000] = qpow(fac[5000], mod - 2);
  for (int i = (4999); i >= (1); i--) inv[i] = inv[i + 1] * (i + 1) % mod;
}
long long getC(int n, int m) {
  if (m < 0 || n < 0 || n - m < 0) return 0;
  return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
long long tmp[5050];
void solve() {
  init();
  long long l, n, k;
  cin >> n >> k >> l;
  for (int i = (k); i <= (n); i++)
    for (int j = (0); j <= (i); j++)
      tmp[i] = (tmp[i] + getC(i, j) * tinv[i + j + 1] % mod * (j % 2 ? -1 : 1) +
                mod) %
               mod;
  long long ans = 0;
  for (int i = (k); i <= (n); i++) {
    long long d = 0;
    for (int j = (0); j <= (n - i); j++)
      d = (d +
           n2[i + j] * getC(n - i, j) % mod * tmp[i + j] % mod *
               (j % 2 ? -1 : 1) +
           mod) %
          mod;
    ans = (ans + d * getC(n, i)) % mod;
  }
  cout << ans * l % mod << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  { solve(); }
  return 0;
}