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 maxn = 4010, mod = (119 << 23) + 1;
int C[maxn], fac[maxn], ifac[maxn], inv[maxn];
inline int _C(int n, int k) {
  if (k > n) return 0;
  return 1ll * fac[n] * ifac[k] % mod * ifac[n - k] % mod;
}
int Pow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod) {
    if (b & 1) ans = 1ll * ans * a % mod;
  }
  return ans;
}
void Mul(int a, int b, int c) {
  for (int i = maxn - 1; i > 1; i--)
    C[i] = (1ll * C[i] * a + 1ll * C[i - 1] * b + 1ll * C[i - 2] * c) % mod;
  C[1] = (1ll * C[1] * a + 1ll * C[0] * b) % mod;
  C[0] = (1ll * C[0] * a) % mod;
}
void Div(int a, int b, int c) {
  int rv = inv[a];
  C[0] = 1ll * C[0] * rv % mod;
  C[1] = 1ll * rv * (1ll * C[1] - 1ll * C[0] * b) % mod;
  for (int i = 2; i < maxn; i++)
    C[i] =
        1ll * rv * (1ll * C[i] - 1ll * C[i - 1] * b - 1ll * C[i - 2] * c) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  fac[0] = 1;
  for (int i = 1; i < maxn; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[maxn - 1] = Pow(fac[maxn - 1], mod - 2);
  for (int i = maxn - 2; i >= 0; i--)
    ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  for (int i = 1; i < maxn; i++) inv[i] = 1ll * ifac[i] * fac[i - 1] % mod;
  int n, k, l, ans = 0;
  cin >> n >> k >> l;
  C[0] = 1;
  for (int i = 0; i < k - 1; i++) Mul(0, 2, -2);
  for (int i = 0; i < n - k + 1; i++) Mul(1, -2, 2);
  for (int i = k; i <= n; i++) {
    Mul(0, 2, -2);
    Div(1, -2, 2);
    for (int j = 0; j <= 2 * n; j++)
      ans = (1ll * ans + 1ll * _C(n, i) * C[j] % mod * inv[j + 1]) % mod;
  }
  ans = 1ll * ans * l % mod;
  if (ans < 0) ans += mod;
  return cout << ans << endl, 0;
}