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 long long MOD = 998244353;
const int N = 2010;
long long inv[N << 1], fac[N], ifac[N], now[N << 1], b[N << 1], n, n2, k, l,
    ans = 0;
long long pw(long long x, int n) {
  long long res = 1;
  for (; n; n >>= 1, x = x * x % MOD)
    if (n & 1) res = res * x % MOD;
  return res;
}
long long C(int a, int b) { return fac[a] * ifac[b] % MOD * ifac[a - b] % MOD; }
void mul1(long long a[], int n) {
  int n2 = n << 1;
  for (int i = 0; i <= n2; ++i) b[i] = a[i];
  a[0] = 0;
  a[1] = (b[0] << 1) % MOD;
  for (int j = 2; j <= n2; ++j)
    a[j] = (-(b[j - 2] << 1) + (b[j - 1] << 1) + (MOD << 2)) % MOD;
}
void mul2(long long a[]) {
  for (int i = 0; i <= n2; ++i) b[i] = a[i];
  a[0] = b[0];
  a[1] = (-(b[0] << 1) + b[1] + (MOD << 2)) % MOD;
  for (int j = 2; j <= n2; ++j)
    a[j] = ((b[j - 2] << 1) - (b[j - 1] << 1) + b[j] + (MOD << 2)) % MOD;
}
void div(long long a[]) {
  memset(b, 0, sizeof b);
  for (int i = 0; i <= n2; ++i) b[i] = MOD - a[i + 1] * inv[2] % MOD;
  for (int i = n2; i >= 0; --i) a[i] = b[i + 1] % MOD, b[i] += b[i + 1];
}
int main() {
  cin >> n >> k >> l;
  n2 = n << 1;
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % MOD;
  ifac[n] = pw(fac[n], MOD - 2);
  for (int i = n - 1; i >= 0; --i) ifac[i] = ifac[i + 1] * (i + 1) % MOD;
  for (int i = 1; i <= n + 1 << 1; ++i) inv[i] = pw(i, MOD - 2);
  now[0] = 1;
  for (int i = 1; i <= n; ++i) mul1(now, i);
  for (int t = n; t >= k; --t) {
    long long anst = 0;
    for (int i = 0; i <= n << 1; ++i) anst = (anst + now[i] * inv[i + 1]) % MOD;
    ans = (ans + anst * C(n, t)) % MOD;
    div(now);
    mul2(now);
  }
  cout << (ans + MOD) % MOD * l % MOD << endl;
}