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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long N = 998244353, M = 1000000;
long long n, m, t, k, k2, f, mi = 2e9, x, y, z, ma, x3, x2, y2, l, r, i, j, ans,
                             b[410000], d[400110], a[410000], dp[2010][2010],
                             dp2[2010][2010], p[410000];
string s, s2;
vector<long long> v[410000], v2, v3;
long long vozv(long long x, long long y) {
  if (y == 0) return 1ll;
  long long t = vozv(x, y / 2);
  if (y % 2) return t * t % N * x % N;
  return t * t % N;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  ans = k * vozv(n * 2 + 1, N - 2) % N;
  l = 0;
  r = 0;
  dp[1][0] = 1;
  if (m == 1) dp2[1][0] = 1;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= i; j++) {
      if (i == 1 && !j) continue;
      if (j != i) dp[i][j] = (dp[i][j] + dp[i - 1][j]) % N;
      if (j) dp[i][j] = (dp[i][j] + dp[i][j - 1] * (i - j + 1)) % N;
      if (j != i) dp2[i][j] = (dp2[i][j] + dp2[i - 1][j]) % N;
      if (j) dp2[i][j] = (dp2[i][j] + dp2[i][j - 1] * (i - j + 1)) % N;
      if (i - j >= m) dp2[i][j] = (dp2[i][j] + dp[i][j]) % N;
    }
  }
  l = dp[n][n];
  r = dp2[n][n];
  ans = ans * r % N * vozv(l, N - 2) % N;
  cout << ans;
}