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 optimize("Ofast")
using namespace std;
const long long inf = 0x3f3f3f3f;
const double eps = 1e-6;
const long long mod = 998244353;
long long modpow(long long x, long long y) {
  long long res = 1, p = x;
  do {
    if (y & 1) res = res * p % mod;
    p = p * p % mod;
  } while (y >>= 1);
  return res;
}
long long n, k, l;
long long dp[4005][2005], c[2005][2005];
signed main() {
  c[0][0] = 1;
  dp[0][0] = 1;
  cin >> n >> k >> l;
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= i; j++)
      if (i || j)
        c[i][j] = ((!i ? 0 : c[i - 1][j] * (i - j)) + (!j ? 0 : c[i][j - 1])),
        c[i][j] %= mod;
  long long ans = 0;
  for (long long i = 0; i <= n + n; i++) {
    for (long long j = 0; j <= n && j <= i && j <= (n + n - i); j++) {
      if (dp[i][j]) {
        long long right = (i - j) / 2, left = (i + j) / 2;
        left = n - left;
        right = n - right;
        long long Left = !left ? 0ll : c[right][left - 1],
                  Right = (left == right) ? 0ll : c[right - 1][left] * j;
        Left %= mod;
        Right %= mod;
        long long Inv = modpow(Left + Right, mod - 2);
        if (j >= k) ans += dp[i][j], ans %= mod;
        dp[i][j] = (dp[i][j] * Inv) % mod;
        (dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j] * (Left) % mod) % mod);
        if (j)
          (dp[i + 1][j - 1] =
               (dp[i + 1][j - 1] + dp[i][j] * (Right) % mod) % mod);
      }
    }
  }
  ans = ans * l % mod * modpow(n + n + 1, mod - 2) % mod;
  cout << ans << endl;
  return 0;
}