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 MAX = 4010, MOD = 998244353;
long long dp[MAX][MAX][2], fact[MAX];
int n, k, l;
void precalc() {
  fact[0] = 1ll;
  for (int i = 1; i < MAX; i++) fact[i] = (1ll * i * fact[i - 1]) % MOD;
}
long long fpow(long long b, int p) {
  long long r = 1ll;
  while (p > 0) {
    if (p & 1) r = (1ll * r * b) % MOD;
    p >>= 1;
    b = (1ll * b * b) % MOD;
  }
  return r;
}
long long solve() {
  long long tmp = (1ll * dp[2 * n + 1][0][1] * fpow(2, n)) % MOD;
  tmp = (1ll * tmp * fact[n]) % MOD;
  tmp = (1ll * tmp * fpow(fact[2 * n + 1], MOD - 2)) % MOD;
  return (1ll * tmp * l) % MOD;
}
int main() {
  scanf("%d%d%d", &n, &k, &l);
  precalc();
  dp[0][0][0] = 1;
  for (int i = 0; i <= 2 * n + 1; i++) {
    for (int j = 0; j <= n; j++) {
      for (int b = 0; b <= 1; b++) {
        if (j >= k && b == 0)
          dp[i + 1][j][1] = (dp[i + 1][j][1] + dp[i][j][b]) % MOD;
        if (j)
          dp[i + 1][j - 1][b] =
              (dp[i + 1][j - 1][b] + 1ll * dp[i][j][b] * j % MOD) % MOD;
        if (i + j + 1 <= 2 * n + b)
          dp[i + 1][j + 1][b] = (dp[i + 1][j + 1][b] + dp[i][j][b]) % MOD;
      }
    }
  }
  printf("%d\n", solve());
  return 0;
}