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 mod = 998244353;
long long int qpow(long long int x, long long int y) {
  long long int ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
const int maxn = 1024 * 2;
long long int dp[maxn * 2][maxn][2];
long long int inv[maxn * 4];
long long int Solve(int n, int k) {
  dp[1][0][1] = 1;
  long long int tem = qpow(2, mod - 2);
  long long int ans = 0;
  for (int i = 1; i < maxn * 4; i++) inv[i] = qpow(i, mod - 2);
  for (int i = 1; i <= 2 * n; i++) {
    for (int j = 0; j <= n; j++) {
      if (j >= k) {
        ans = (ans + dp[i][j][0]) % mod;
        ans = (ans + dp[i][j][1]) % mod;
      }
      if (j - 1 <= (2 * n - i) && j >= 1) {
        if (j - 1 == 2 * n - i) {
          dp[i + 1][j - 1][0] = (dp[i + 1][j - 1][0] + dp[i][j][0]) % mod;
        } else if (j == 1) {
          dp[i + 1][j - 1][1] = (dp[i + 1][j - 1][1] + dp[i][j][0]) % mod;
        } else {
          long long int send = (2 * n - (i + j - 1));
          long long int reci = j - 1;
          long long int invt = inv[send + reci];
          dp[i + 1][j - 1][1] =
              (dp[i + 1][j - 1][1] + dp[i][j][0] * send % mod * invt % mod) %
              mod;
          dp[i + 1][j - 1][0] =
              (dp[i + 1][j - 1][0] + dp[i][j][0] * reci % mod * invt % mod) %
              mod;
        }
      }
      if (j + 1 <= (2 * n - i)) {
        if (j + 1 == (2 * n - i)) {
          dp[i + 1][j + 1][0] = (dp[i + 1][j + 1][0] + dp[i][j][1]) % mod;
        } else {
          long long int send = (2 * n - (i + j + 1));
          long long int reci = j + 1;
          long long int invt = inv[send + reci];
          dp[i + 1][j + 1][1] =
              (dp[i + 1][j + 1][1] + dp[i][j][1] * send % mod * invt % mod) %
              mod;
          dp[i + 1][j + 1][0] =
              (dp[i + 1][j + 1][0] + dp[i][j][1] * reci % mod * invt % mod) %
              mod;
        }
      }
    }
  }
  return ans;
}
int main() {
  int n, k, l;
  scanf("%d%d%d", &n, &k, &l);
  long long int ans = Solve(n, k) * qpow(2 * n + 1, mod - 2) % mod * l % mod;
  printf("%lld", ans);
}