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;
int n, K, L = 4007;
long long T[4010], T2[4010], Mod = 998244353, Inv2 = Mod / 2 + 1,
                             Inv4 = Inv2 * Inv2 % Mod, Len, RR[4010];
long long A[2010][4010], B[2010][4010],
    PA[3] = {0, 2, Mod - 2}, PB[3] = {1, Mod - 2, 2}, Comb[2020][2020];
long long Inv(long long a) {
  long long b = Mod - 2, r = 1;
  while (b) {
    if (b & 1) r = r * a % Mod;
    a = a * a % Mod;
    b >>= 1;
  }
  return r;
}
int main() {
  int i, j, k;
  scanf("%d%d%lld", &n, &K, &Len);
  A[0][0] = 1;
  B[0][0] = 1;
  for (i = 0; i <= 2000; i++) {
    Comb[i][0] = 1;
    for (j = 1; j <= i; j++)
      Comb[i][j] = (Comb[i - 1][j - 1] + Comb[i - 1][j]) % Mod;
  }
  for (i = 1; i <= n + 1; i++) {
    for (j = 0; j < 2 * i; j++) {
      for (k = 0; k < 3; k++) {
        A[i][j + k] = (A[i][j + k] + 1ll * A[i - 1][j] * PA[k]) % Mod;
        B[i][j + k] = (B[i][j + k] + 1ll * B[i - 1][j] * PB[k]) % Mod;
      }
    }
  }
  for (i = 0; i <= L; i++) T[i] = A[n][i];
  for (i = n; i >= K; i--) {
    for (j = 0; j <= L; j++) {
      RR[j] = (RR[j] + T[j] * Comb[n][i]) % Mod;
      T2[j] = 0;
    }
    for (j = L; j >= 0; j--) {
      if (T[j]) {
        long long t = T[j] * Inv2 % Mod;
        t = Mod - t;
        T2[j - 2] = t;
        T[j] = (T[j] + 2 * t) % Mod;
        T[j - 1] = (T[j - 1] - 2 * t + Mod * 2) % Mod;
      }
    }
    for (j = 0; j <= L; j++) T[j] = 0;
    for (j = 0; j <= L - 2; j++) {
      for (k = 0; k < 3; k++) {
        T[j + k] = (T[j + k] + T2[j] * PB[k]) % Mod;
      }
    }
  }
  long long res = 0;
  for (i = 0; i <= L; i++) {
    res = (res + RR[i] * Inv(i + 1)) % Mod;
  }
  printf("%lld\n", res * Len % Mod);
}