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 M = 998244353;
const long long MN = 2020;
inline long long mul(long long a, long long b) { return a * b % M; }
long long bp(long long b, long long p) {
  long long ac = 1;
  while (p) {
    if (p & 1) {
      ac = mul(ac, b);
    }
    b = mul(b, b);
    p >>= 1;
  }
  return ac;
}
inline long long norm(long long b) { return ((b % M) + M) % M; }
long long di[MN * 3];
long long inv(long long b) {
  if (b < MN * 3) {
    if (di[b] != -1) return di[b];
    return di[b] = bp(b, M - 2);
  }
  return bp(b, M - 2);
}
long long dp[MN][MN];
long long ch(long long n, long long k) {
  if (k < 0 || k > n) return 0;
  if ((k == 0) || (k == n)) return 1;
  if (dp[n][k] != -1) return dp[n][k];
  long long res = ch(n - 1, k - 1) + ch(n - 1, k);
  if (res >= M) {
    res -= M;
  }
  return dp[n][k] = res;
}
long long dq[MN];
long long calc(long long p) {
  if (dq[p] != -1) return dq[p];
  long long res = 0;
  long long sn = 1;
  for (int i = 0; i <= p; i++) {
    res += (ch(p, i) * sn) % M * inv(p + i + 1) % M;
    sn = M - sn;
  }
  res = mul(norm(res), bp(2, p));
  return dq[p] = res;
}
long long get(long long n, long long k) {
  long long res = 0;
  long long sn = 1;
  for (int i = k; i <= n; i++) {
    res += (ch(n - k, i - k) * sn) % M * calc(i) % M;
    sn = M - sn;
  }
  return norm(res);
}
int main() {
  long long n, k, l;
  cin >> n >> k >> l;
  memset(dq, -1, sizeof(dq));
  memset(dp, -1, sizeof(dp));
  memset(di, -1, sizeof(di));
  long long res = 0;
  for (int i = k; i <= n; i++) {
    res += mul(get(n, i), ch(n, i));
  }
  res = mul(norm(res), norm(l));
  cout << norm(res) << '\n';
}