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 llINF = 9223372036854775807;
const int INF = 2147483647;
const int maxn = 3e5 + 7;
const int maxm = 5e4 + 7;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-12;
const int N = 100005;
const int M = 998244353;
int n, i, l, k;
long long dp[2005][2005], f[2005][2005];
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long g = exgcd(b, a % b, x, y);
  long long tmp = x;
  x = y;
  y = tmp - a / b * y;
  return g;
}
long long ni(long long x) {
  long long a, b;
  exgcd(x, M, a, b);
  return a;
}
long long dfs2(int i, int j) {
  if (i == n && j == 0) return f[i][j] = 1;
  if (i > n) return 0;
  if (f[i][j] != -1) return f[i][j];
  if (j != 0)
    return f[i][j] = (dfs2(i, j - 1) * j + dfs2(i + 1, j + 1)) % M;
  else
    return f[i][j] = dfs2(i + 1, j + 1) % M;
}
long long dfs(int i, int j) {
  if (i == n && j == 0) return 0;
  if (i > n) return 0;
  if (dp[i][j] != -1) return dp[i][j];
  if (j >= k)
    return dp[i][j] = (dfs(i + 1, j + 1) + dfs(i, j - 1) * j + f[i][j]) % M;
  else if (j != 0)
    return dp[i][j] = (dfs(i, j - 1) * j + dfs(i + 1, j + 1)) % M;
  else
    return dp[i][j] = dfs(i + 1, j + 1) % M;
}
int main() {
  scanf("%d %d %d", &n, &k, &l);
  memset(dp, -1, sizeof(dp));
  memset(f, -1, sizeof(f));
  dfs2(0, 0);
  cout << (dfs(0, 0) * ni(f[0][0]) % M * ni(2 * n + 1) % M * l % M + M) % M;
}