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 N = 2e3 + 10, mod = 998244353;
long long choose[N][N];
long long pre_inverse[2 * N];
void polymul(vector<long long> &a, vector<long long> b);
void polydiv(vector<long long> &a, long long l);
long long mul(long long x, long long y);
long long inverse(long long a);
long long pwr(long long x, long long e);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  long long l;
  cin >> n >> k >> l;
  for (int i = 0; i < N; ++i) {
    choose[i][0] = choose[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      choose[i][j] = (choose[i - 1][j - 1] + choose[i - 1][j]) % mod;
    }
  }
  for (int i = 1; i < 2 * N; ++i) {
    pre_inverse[i] = inverse(i);
  }
  vector<long long> p = {0, mul(2, inverse(l)),
                         mul(mod - 2, inverse(mul(l, l)))};
  vector<long long> q = p;
  for (auto &i : q) {
    i = (mod - i) % mod;
  }
  q[0] = 1;
  vector<long long> P = {1};
  for (int i = 0; i < n; ++i) {
    polymul(P, p);
  }
  long long ans = 0;
  for (int i = n; i >= k; --i) {
    int deg = P.size();
    long long temp, l_pwr = 1;
    for (int j = 0; j < deg; ++j) {
      l_pwr = mul(l_pwr, l);
      temp = mul(P[j], pre_inverse[j + 1]);
      temp = mul(temp, choose[n][i]);
      temp = mul(temp, l_pwr);
      ans += temp;
    }
    ans %= mod;
    polymul(P, q);
    polydiv(P, l);
  }
  cout << ans % mod << endl;
}
void polymul(vector<long long> &a, vector<long long> b) {
  int d1 = a.size(), d2 = b.size();
  vector<long long> ans(d1 + d2 - 1, 0);
  for (int i = 0; i < d1; ++i) {
    for (int j = 0; j < d2; ++j) {
      ans[i + j] += a[i] * b[j];
      ans[i + j] %= mod;
    }
  }
  a = ans;
}
void polydiv(vector<long long> &a, long long l) {
  int d = a.size();
  vector<long long> ans(d - 1);
  for (int i = d - 1; i > 0; --i) {
    long long temp = mod - a[i];
    temp %= mod;
    ans[i - 1] = temp;
    a[i - 1] -= mul(temp, l);
    a[i - 1] += mod;
    a[i - 1] %= mod;
  }
  for (int i = 0; i < d - 1; ++i) {
    ans[i] = mul(ans[i + 1], pre_inverse[2]);
    ans[i] = mul(ans[i], mul(l, l));
  }
  ans.pop_back();
  a = ans;
}
long long mul(long long x, long long y) {
  long long ans = x * y;
  ans %= mod;
  if (ans < 0) {
    ans += mod;
  }
  return ans;
}
long long inverse(long long a) { return pwr(a, mod - 2); }
long long pwr(long long x, long long e) {
  long long res = 1;
  while (e) {
    if (e & 1) {
      res = (res * x) % mod;
    }
    e /= 2;
    x = (x * x) % mod;
  }
  return res;
}