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 mod = 998244353;
const int maxn = 4001;
int n, l, k;
int dp[2005][2005][2];
long long fir[4005], inv[4005];
int adds(int v) { return (v >= mod) ? v - mod : v; }
long long ksm(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ans;
}
void pre() {
  fir[0] = 1;
  for (int i = 1; i <= maxn; i++) fir[i] = fir[i - 1] * i % mod;
  inv[maxn] = ksm(fir[maxn], mod - 2) % mod;
  for (int i = maxn - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
}
long long C(int n, int m) { return fir[n] * inv[m] % mod * inv[n - m] % mod; }
int main() {
  pre();
  scanf("%d%d%d", &n, &k, &l);
  dp[0][0][0] = 1;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= i; j++)
      for (int op = 0; op <= 1; op++) {
        if (!dp[i][j][op]) continue;
        if (j < i)
          dp[i][j + 1][op] =
              adds(dp[i][j + 1][op] + 1ll * dp[i][j][op] * (i - j) % mod);
        if (i < n) dp[i + 1][j][op] = adds(dp[i + 1][j][op] + dp[i][j][op]);
        if (!op && i - j >= k)
          dp[i][j][op + 1] = adds(dp[i][j][op + 1] + dp[i][j][op]);
      }
  long long ans = dp[n][n][1] * ksm(2, n) % mod * fir[n] % mod *
                  inv[2 * n + 1] % mod * l % mod;
  printf("%lld\n", ans);
}