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 INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 998244353;
const long long maxn = 1e6 + 107;
const double pi = acos(-1.0);
const double eps = 0.0000000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline void pr2(T x, int k = 64) {
  long long i;
  for (i = 0; i < k; i++) fprintf(stderr, "%d", (x >> i) & 1);
  putchar(' ');
}
template <typename T>
inline void add_(T &A, int B, long long MOD = M) {
  A += B;
  (A >= MOD) && (A -= MOD);
}
template <typename T>
inline void mul_(T &A, long long B, long long MOD = M) {
  A = (A * B) % MOD;
}
template <typename T>
inline void mod_(T &A, long long MOD = M) {
  A %= MOD;
  A += MOD;
  A %= MOD;
}
template <typename T>
inline void max_(T &A, T B) {
  (A < B) && (A = B);
}
template <typename T>
inline void min_(T &A, T B) {
  (A > B) && (A = B);
}
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
inline long long powMM(long long a, long long b, long long mod = M) {
  long long ret = 1;
  for (; b; b >>= 1ll, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
int startTime;
void startTimer() { startTime = clock(); }
void printTimer() {
  fprintf(stderr, "/--- Time: %ld milliseconds ---/\n", clock() - startTime);
}
int f[5007], g[5007], val[5007], mul[5007];
long long inv[1000002];
long long Inv[1000002];
long long fac[1000002];
long long C(int n, int m) { return fac[n] * inv[m] % M * inv[n - m] % M; }
int main() {
  int N, K, L, i, k;
  scanf("%d%d%d", &N, &K, &L);
  fac[0] = 1;
  for (i = 1; i <= 1000000; i++) fac[i] = i * fac[i - 1] % M;
  inv[0] = inv[1] = 1;
  for (i = 2; i <= 1000000; i++) inv[i] = (M - M / i) * inv[M % i] % M;
  for (i = 1; i <= 1000000; i++)
    Inv[i] = inv[i], inv[i] = inv[i] * inv[i - 1] % M;
  f[0] = 1;
  for (i = 1; i <= N; i++) {
    for (k = 0; k <= N * 2; k++) g[k] = f[k], f[k] = 0;
    for (k = 0; k <= N * 2; k++) {
      add_(f[k + 1], (long long)2 * g[k] % M);
      add_(f[k + 2], M - (long long)2 * g[k] % M);
    }
  }
  int ans = 0;
  for (i = N; i >= K; i--) {
    int now = 0;
    for (k = 0; k <= N * 2; k++) add_(now, (long long)f[k] * Inv[k + 1] % M);
    add_(ans, (long long)C(N, i) * now % M);
    for (k = 0; k <= N * 2; k++) add_(f[k + 1], f[k]);
    for (k = 0; k <= N * 2; k++) f[k] = (long long)f[k + 1] * Inv[2] % M;
    for (k = 0; k <= N * 2; k++) g[k] = f[k], f[k] = 0;
    for (k = 0; k <= N * 2; k++) {
      add_(f[k], g[k]);
      add_(f[k + 1], M - (long long)2 * g[k] % M);
      add_(f[k + 2], (long long)2 * g[k] % M);
    }
  }
  mul_(ans, L);
  printf("%d\n", ans);
}