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>
int mod = 998244353;
long long int cres[2010][4010];
long long int tmp[4010];
long long int fct[4010];
int n, m, p;
void lpow(int x) {
  for (int i = 0; i <= 2 * n; i++) {
    tmp[i] = 0;
  }
  for (int i = 0; i <= 2 * n; i++) {
    tmp[i] += cres[x][i];
    tmp[i + 1] += 2 * (mod - cres[x][i]);
    tmp[i + 2] += 2 * cres[x][i];
  }
  for (int i = 0; i <= 2 * n; i++) {
    cres[x][i] = tmp[i] % mod;
  }
}
void rpow(int x) {
  for (int i = 0; i <= 2 * n; i++) {
    tmp[i] = 0;
  }
  for (int i = 0; i <= 2 * n; i++) {
    tmp[i + 1] += 2 * cres[x][i];
    tmp[i + 2] += 2 * (mod - cres[x][i]);
  }
  for (int i = 0; i <= 2 * n; i++) {
    cres[x][i] = tmp[i] % mod;
  }
}
void pcopy(int x, int y) {
  for (int i = 0; i <= 2 * n; i++) {
    cres[y][i] = cres[x][i];
  }
}
void divconq(int l, int r) {
  if (l == r) return;
  int lmid = (l + r) / 2;
  int rmid = lmid + 1;
  pcopy(l, rmid);
  for (int i = 0; i < (r - lmid); i++) {
    lpow(l);
  }
  for (int i = 0; i < (rmid - l); i++) {
    rpow(rmid);
  }
  divconq(l, lmid);
  divconq(rmid, r);
}
long long int mpow(int x, int y) {
  if (y == 0) return 1;
  long long int r = mpow(x, y / 2);
  r *= r;
  r %= mod;
  if (y % 2 == 1) {
    r *= x;
    r %= mod;
  }
  return r;
}
long long int minv(int x) { return mpow(x, mod - 2); }
long long int comb(int x, int y) {
  return fct[x] * minv(fct[y]) % mod * minv(fct[x - y]) % mod;
}
int main() {
  scanf("%d%d%d", &n, &m, &p);
  cres[0][0] = 1;
  divconq(0, n);
  long long int ans = 0;
  fct[0] = 1;
  for (int i = 0; i <= 2 * n; i++) {
    tmp[i] = minv(i + 1);
    fct[i + 1] = (fct[i] * (i + 1)) % mod;
  }
  for (int i = m; i <= n; i++) {
    long long int tans = 0;
    for (int j = 0; j <= 2 * n; j++) {
      tans += tmp[j] * cres[i][j];
      tans %= mod;
    }
    tans *= comb(n, i);
    tans %= mod;
    ans += tans;
    ans %= mod;
  }
  ans *= p;
  ans %= mod;
  printf("%lld", ans);
  return 0;
}