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;
inline void read(int &x) {
  x = 0;
  int p = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') p = -p;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch - '0'), ch = getchar();
  x = x * p;
}
int l, n, k, c[4005][4005], inv[4005], f[2005], po[4005];
int qpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1LL * res * x % mod;
    x = 1LL * x * x % mod;
    y >>= 1;
  }
  return res;
}
inline void add(int &x, int y) { x + y >= mod ? x = x + y - mod : x += y; }
int main() {
  read(n);
  read(k);
  read(l);
  c[0][0] = 1;
  for (int i = 1; i <= 4001; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      c[i][j] >= mod ? c[i][j] -= mod : c[i][j];
    }
  }
  po[0] = 1;
  for (int i = 1; i <= 4001; ++i)
    inv[i] = qpow(i, mod - 2), po[i] = 1LL * po[i - 1] * 2 % mod;
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= i; ++j) {
      add(f[i],
          1LL * c[i][j] * (j & 1 ? mod - 1 : 1) % mod * inv[i + j + 1] % mod);
    }
  }
  int ans = 0;
  for (int m = k; m <= n; ++m) {
    int now = 0;
    for (int j = 0; j <= n - m; ++j) {
      add(now, 1LL * (j & 1 ? mod - 1 : 1) * c[n - m][j] % mod * po[m + j] %
                   mod * f[m + j] % mod);
    }
    now = 1LL * now * c[n][m] % mod;
    add(ans, now);
  }
  ans = 1LL * ans * l % mod;
  cout << ans << endl;
}