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;
long long quick_pow(long long x, long long a) {
  long long ans = 1;
  while (a) {
    if (a & 1) ans = ans * x % MOD;
    x = x * x % MOD;
    a >>= 1;
  }
  return ans;
}
long long f[5005], pow1[2005][5005], pow2[2005][5005], c[2005][2005];
int n, k, l;
int limit, L, r[10005];
long long omega[2][15][1 << 14];
void pre() {
  for (int mid = 1, i = 0; i <= 14; mid <<= 1, i++) {
    long long w1 = quick_pow(3, (MOD - 1) / (mid << 1));
    long long w2 = quick_pow(3, MOD - 1 - (MOD - 1) / (mid << 1));
    omega[0][i][0] = omega[1][i][0] = 1;
    for (int k = 1; k < mid; k++) {
      omega[1][i][k] = omega[1][i][k - 1] * w1 % MOD;
      omega[0][i][k] = omega[0][i][k - 1] * w2 % MOD;
    }
  }
}
void NTT(long long* A, int type) {
  for (int i = 0; i < limit; i++)
    if (i < r[i]) swap(A[i], A[r[i]]);
  if (type == -1) type = 0;
  for (int mid = 1, i = 0; mid < limit; mid <<= 1, i++) {
    for (int R = mid << 1, j = 0; j < limit; j += R) {
      for (int k = 0; k < mid; k++) {
        long long x = A[j + k] % MOD,
                  y = omega[type][i][k] * A[j + mid + k] % MOD;
        A[j + k] = x + y;
        A[j + mid + k] = x - y;
      }
    }
  }
  for (int i = 0; i < limit; i++) {
    A[i] %= MOD;
    if (A[i] < 0) A[i] += MOD;
  }
  if (type == 0) {
    long long inv = quick_pow(limit, MOD - 2);
    for (int i = 0; i < limit; i++) A[i] = A[i] * inv % MOD;
  }
}
long long tmp1[10005], tmp2[10005];
int main() {
  pre();
  scanf("%d%d%d", &n, &k, &l);
  limit = 1, L = 0;
  while (limit <= (2 * n)) limit <<= 1, L++;
  for (int i = 0; i < limit; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));
  c[0][0] = 1;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) {
      if (c[i][j] >= MOD) c[i][j] -= MOD;
      c[i + 1][j + 1] += c[i][j];
      c[i + 1][j] += c[i][j];
    }
  pow1[0][0] = 1;
  pow1[1][1] = 2, pow1[1][2] = -2;
  NTT(pow1[0], 1);
  NTT(pow1[1], 1);
  for (int i = 2; i <= n; i++)
    for (int x = 0; x < limit; x++)
      pow1[i][x] = pow1[i - 1][x] * pow1[1][x] % MOD;
  pow2[0][0] = 1;
  pow2[1][0] = 1, pow2[1][1] = -2, pow2[1][2] = 2;
  NTT(pow2[0], 1);
  NTT(pow2[1], 1);
  for (int i = 2; i <= n; i++)
    for (int x = 0; x < limit; x++)
      pow2[i][x] = pow2[i - 1][x] * pow2[1][x] % MOD;
  for (int i = k; i <= n; i++) {
    for (int j = 0; j < limit; j++) {
      tmp1[j] = pow1[i][j];
      tmp2[j] = pow2[n - i][j];
    }
    for (int j = 0; j < limit; j++) tmp1[j] = tmp1[j] * tmp2[j] % MOD;
    for (int j = 0; j < limit; j++) f[j] = (f[j] + tmp1[j] * c[n][i]) % MOD;
  }
  NTT(f, -1);
  long long ans = 0;
  for (int i = 0; i <= 2 * n; i++)
    ans += quick_pow(i + 1, MOD - 2) * f[i] % MOD;
  ans %= MOD;
  if (ans < 0) ans += MOD;
  printf("%I64d\n", ans * l % MOD);
  return 0;
}