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 N = 2010;
const int MOD = 998244353;
inline int add(int u, int v) {
  u += v;
  if (u >= MOD) u -= MOD;
  return u;
}
inline int sub(int u, int v) {
  u -= v;
  if (u < 0) u += MOD;
  return u;
}
inline int mul(int u, int v) { return (long long)u * v % MOD; }
inline int power(int u, int v) {
  int res = 1;
  while (v) {
    if (v & 1) res = mul(res, u);
    u = mul(u, u);
    v >>= 1;
  }
  return res;
}
inline int inv(int u) { return power(u, MOD - 2); }
int n, k, l;
int a[N], b[N << 1];
int c[N][N];
int pow2[N];
int inverse[N << 1];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> l;
  pow2[0] = 1;
  for (int i = 1; i <= n; i++) {
    pow2[i] = mul(pow2[i - 1], 2);
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = add(c[i - 1][j - 1], c[i - 1][j]);
    }
  }
  for (int i = k; i <= n; i++) {
    for (int j = 0; j <= n - i; j++) {
      int cost = mul(c[n][i], c[n - i][j]);
      if (j & 1) cost = sub(0, cost);
      a[i + j] = add(a[i + j], cost);
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      int cost = mul(a[i], pow2[i]);
      cost = mul(cost, c[i][j]);
      if (j & 1) cost = sub(0, cost);
      b[i + j] = add(b[i + j], cost);
    }
  }
  inverse[1] = 1;
  for (int i = 2; i <= n * 2 + 1; i++) {
    inverse[i] = sub(0, mul(MOD / i, inverse[MOD % i]));
  }
  int res = 0;
  for (int i = 0; i <= n * 2; i++) {
    res = add(res, mul(b[i], inverse[i + 1]));
  }
  res = mul(res, l);
  cout << res << endl;
  return 0;
}