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;
template <typename T1, typename T2>
inline T1 max(T1 a, T2 b) {
  return a < b ? b : a;
}
template <typename T1, typename T2>
inline T1 min(T1 a, T2 b) {
  return a < b ? a : b;
}
const char lf = '\n';
namespace ae86 {
const int bufl = 1 << 15;
char buf[bufl], *s = buf, *t = buf;
inline int fetch() {
  if (s == t) {
    t = (s = buf) + fread(buf, 1, bufl, stdin);
    if (s == t) return EOF;
  }
  return *s++;
}
inline int ty() {
  int a = 0;
  int b = 1, c = fetch();
  while (!isdigit(c)) b ^= c == '-', c = fetch();
  while (isdigit(c)) a = a * 10 + c - 48, c = fetch();
  return b ? a : -a;
}
}  // namespace ae86
using ae86::ty;
const int _ = 2007, __ = _ << 1, mo = 998244353;
template <typename T1, typename T2>
inline T1 ad(T1 &a, T2 b) {
  return a = a + b >= mo ? a + b - mo : a + b;
}
template <typename T1, typename T2>
inline T1 dl(T1 &a, T2 b) {
  return a = a >= b ? a - b : a - b + mo;
}
template <typename T1, typename T2>
inline T1 add(T1 a, T2 b) {
  return a + b >= mo ? a + b - mo : a + b;
}
template <typename T1, typename T2>
inline T1 del(T1 a, T2 b) {
  return a >= b ? a - b : a - b + mo;
}
long long powa(long long a, long long t) {
  long long b = 1;
  while (t) {
    if (t & 1) b = b * a % mo;
    a = a * a % mo, t >>= 1;
  }
  return b;
}
inline long long inva(long long a) { return powa(a, mo - 2); }
long long jc[__], rjc[__];
void fuck(int n = __ - 1) {
  jc[0] = jc[1] = rjc[0] = 1;
  for (int i = 2; i <= n; i++) jc[i] = jc[i - 1] * i % mo;
  rjc[n] = inva(jc[n]);
  for (int i = n - 1; i >= 1; i--) rjc[i] = rjc[i + 1] * (1ll + i) % mo;
}
long long C(int n, int m) {
  if (n < m || n < 0 || m < 0) return 0;
  return jc[n] * rjc[m] % mo * rjc[n - m] % mo;
}
int n, nn, tar, len;
long long f[__][__][2] = {0};
int main() {
  ios::sync_with_stdio(0), cout.tie(nullptr);
  fuck();
  n = ty(), nn = n + n, tar = ty(), len = ty();
  f[0][0][0] = 1;
  for (int i = 0; i <= nn; i++)
    for (int j = 0; j <= i; j++)
      for (int k = 0; k < 2; k++) {
        ad(f[i + 1][j + 1][k], f[i][j][k]);
        if (k <= 0 && j >= tar) ad(f[i + 1][j][1], f[i][j][k]);
        if (j > 0) ad(f[i + 1][j - 1][k], f[i][j][k] * j % mo);
      }
  cout << f[nn + 1][0][1] * powa(2, n) % mo * jc[n] % mo * rjc[nn + 1] % mo *
              len % mo
       << lf;
  return 0;
}