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 MOD = 998244353;
const int N = 4002;
long long n, k, l;
long long inv[N], c[N][N];
long long mpow(long long a, long long x) {
  if (x == 0) return 1;
  long long t = mpow(a, x >> 1);
  if (x % 2 == 0) return t * t % MOD;
  return t * t % MOD * a % MOD;
}
void mul(vector<long long> A, vector<long long> B, vector<long long> &C) {
  C.resize(A.size() + B.size() - 1);
  for (int i = 0; i < C.size(); i++) C[i] = 0;
  for (long long i = 0; i < A.size(); i++) {
    for (long long j = 0; j < B.size(); j++) {
      (C[i + j] += A[i] * B[j]) %= MOD;
    }
  }
}
void prt(vector<long long> A) {
  for (int i = 0; i < A.size(); i++) cout << A[i] << " ";
  cout << endl;
}
void divide(vector<long long> &A) {
  vector<long long> C, D;
  for (int i = 0; i < A.size(); i++) A[i] = (A[i] * (MOD + 1) / 2) % MOD;
  for (int i = 0; i < (int)A.size() - 1; i++) C.push_back(A[i + 1]);
  D.push_back(C[0]);
  for (int i = 1; i < (int)C.size() - 1; i++) {
    D.push_back((C[i] + D[(int)D.size() - 1]) % MOD);
  }
  A = D;
}
long long Int(vector<long long> A) {
  long long ans = 0;
  for (long long i = 0; i < A.size(); i++) {
    ans = (ans + A[i] * inv[i + 1] % MOD) % MOD;
  }
  return ans;
}
vector<long long> p1[N], p2[N];
int main() {
  inv[0] = 1;
  c[0][0] = 1;
  for (long long i = 1; i < N; i++) c[i][0] = 1, inv[i] = mpow(i, MOD - 2);
  for (long long i = 1; i < N; i++) {
    for (long long j = 1; j <= i; j++) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
    }
  }
  cin >> n >> k >> l;
  p1[0].push_back(1);
  p2[0].push_back(1);
  p1[1].push_back(0);
  p1[1].push_back(2);
  p1[1].push_back(MOD - 2);
  p2[1].push_back(1);
  p2[1].push_back(MOD - 2);
  p2[1].push_back(2);
  for (long long i = 2; i <= n; i++) {
    mul(p1[i - 1], p1[1], p1[i]);
  }
  vector<long long> ret = p1[n];
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    if (n - i + 1 >= k) ans = (ans + Int(ret) * c[n][n - i + 1]) % MOD;
    mul(ret, p2[1], ret);
    divide(ret);
  }
  cout << ans * l % MOD << endl;
}