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;
template <typename TYPE>
inline void chkmax(TYPE &first, TYPE second) {
  first < second ? first = second : TYPE();
}
template <typename TYPE>
inline void chkmin(TYPE &first, TYPE second) {
  second < first ? first = second : TYPE();
}
template <typename TYPE>
void readint(TYPE &first) {
  first = 0;
  int f = 1;
  char c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) first = first * 10 + c - '0';
  first *= f;
}
const int MAXN = 4005, MOD = 998244353;
inline int dmy(int first) { return first >= MOD ? first - MOD : first; }
inline void inc(int &first, int second) { first = dmy(first + second); }
inline int qmi(int first, int second) {
  int ans = 1;
  for (; second; second >>= 1, first = 1ll * first * first % MOD)
    if (second & 1) ans = 1ll * ans * first % MOD;
  return ans;
}
int n, m, l, a[MAXN], res[MAXN], len = 0, temp[MAXN], c[MAXN][MAXN];
void mul(int A, int B, int C) {
  inc(A, MOD), inc(B, MOD), inc(C, MOD);
  memset(temp, 0, sizeof(temp));
  for (int i = 0; i <= len; ++i)
    inc(temp[i], 1ll * a[i] * C % MOD), inc(temp[i + 1], 1ll * a[i] * B % MOD),
        inc(temp[i + 2], 1ll * a[i] * A % MOD);
  len += 2;
  memcpy(a, temp, sizeof(a));
}
void div(int B) {
  inc(B, MOD);
  memset(temp, 0, sizeof(temp));
  for (int i = len; i; --i)
    inc(temp[i - 1], a[i]), inc(a[i - 1], MOD - 1ll * a[i] * B % MOD);
  --len;
  memcpy(a, temp, sizeof(a));
}
void add(int B) {
  for (int i = 0; i <= 2 * n; ++i) inc(res[i], 1ll * a[i] * B % MOD);
}
int main() {
  readint(n), readint(m), readint(l);
  a[0] = 1;
  for (int i = 1; i <= n; ++i) mul(-2, 2, 0);
  for (int i = 0; i <= n; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= i; ++j) c[i][j] = dmy(c[i - 1][j - 1] + c[i - 1][j]);
  }
  for (int i = n; i >= m; --i) {
    add(c[n][i]);
    div(-1), div(0);
    mul(-1, 1, -qmi(2, MOD - 2));
  }
  int ans = 0;
  for (int i = 0; i <= 2 * n; ++i)
    inc(ans, 1ll * res[i] * qmi(i + 1, MOD - 2) % MOD);
  printf("%I64d\n", 1ll * ans * l % MOD);
  return 0;
}