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;
void getint(int &v) {
  char ch, fu = 0;
  for (ch = '*'; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') fu = 1, ch = getchar();
  for (v = 0; ch >= '0' && ch <= '9'; ch = getchar()) v = v * 10 + ch - '0';
  if (fu) v = -v;
}
const int MO = 998244353;
vector<long long> P, Q;
long long ans, xs[100010], fac[100010], inv[100010], INV[100010];
int n, k, l;
vector<long long> mul(vector<long long> A, vector<long long> B) {
  vector<long long> C;
  C.clear();
  C.resize(((int)A.size()) + ((int)B.size()) - 1);
  for (int i = 0; i <= ((int)A.size()) - 1; ++i)
    for (int j = 0; j <= ((int)B.size()) - 1; ++j)
      (C[i + j] += A[i] * B[j]) %= MO;
  return C;
}
long long pw(long long x, long long y) {
  long long t = 1;
  for (; y; y >>= 1) {
    if (y & 1) t = t * x % MO;
    x = x * x % MO;
  }
  return t;
}
long long C(int x, int y) { return fac[x] * inv[y] % MO * inv[x - y] % MO; }
int main() {
  cin >> n >> k >> l;
  fac[0] = 1;
  for (int i = 1; i <= 100000; ++i) fac[i] = fac[i - 1] * i % MO;
  inv[100000] = pw(fac[100000], MO - 2);
  for (int i = 99999; i >= 0; --i) inv[i] = inv[i + 1] * (i + 1) % MO;
  for (int i = 1; i <= 100000; ++i) INV[i] = pw(i, MO - 2);
  xs[k] = 1;
  for (int i = k + 1; i <= n; ++i) {
    xs[i] = 0;
    for (int j = k; j <= i - 1; ++j) (xs[i] += xs[j] * C(i, j)) %= MO;
    xs[i] = (1 - xs[i]) % MO;
  }
  P.clear();
  P.push_back(1);
  Q.clear();
  Q.push_back(0);
  Q.push_back(2);
  Q.push_back(MO - 2);
  for (int i = 1; i <= k - 1; ++i) P = mul(P, Q);
  for (int i = k; i <= n; ++i) {
    long long t = xs[i] * C(n, i) % MO;
    P = mul(P, Q);
    for (int j = 0; j <= ((int)P.size()) - 1; ++j) {
      (ans += P[j] * INV[j + 1] % MO * t) %= MO;
    }
  }
  ans = ans * l % MO;
  ans = (ans % MO + MO) % MO;
  cout << ans << endl;
  return 0;
}