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;
inline long long read() {
  long long x = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * neg;
}
inline void print(long long x) {
  if (x < 0) {
    putchar('-');
    print(abs(x));
    return;
  }
  if (x <= 9)
    putchar(x + '0');
  else {
    print(x / 10);
    putchar(x % 10 + '0');
  }
}
const long long MOD = 998244353;
inline long long qpow(long long x, long long e, long long _MOD) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * x % _MOD;
    x = x * x % _MOD;
    e >>= 1;
  }
  return ans;
}
long long n = read(), k = read(), l = read();
long long dp[4005][2005], f[4005];
signed main() {
  f[0] = 1;
  for (long long i = 1; i <= n << 1; i++) f[i] = f[i - 1] * i % MOD;
  dp[0][0] = 1;
  for (long long i = 1; i <= (n * 2); i++) {
    for (long long j = 0; j <= min(n, i); j++) {
      if (j) dp[i][j] = dp[i - 1][j - 1];
      dp[i][j] = (dp[i][j] + dp[i - 1][j + 1] * (j + 1) % MOD) % MOD;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n << 1; ++i)
    for (long long j = k; j <= n; ++j)
      ans = (ans + 1ll * dp[i][j] * dp[(n << 1) - i][j] % MOD * f[j]) % MOD;
  ans = ans * qpow(dp[n << 1][0], MOD - 2, MOD) % MOD *
        qpow(2 * n + 1, MOD - 2, MOD) % MOD * l % MOD;
  cout << ans << endl;
  return 0;
}