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;
using LL = long long;
namespace _buff {
const size_t BUFF = 1 << 19;
char buff[BUFF], *begin = buff, *end = buff;
char getc() {
  if (begin == end) {
    begin = buff;
    end = buff + fread(buff, 1, BUFF, stdin);
  }
  return begin == end ? -1 : *begin++;
}
}  // namespace _buff
LL read() {
  using namespace _buff;
  LL ret = 0;
  bool pos = true;
  char c = getc();
  for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
    assert(~c);
  }
  if (c == '-') {
    pos = false;
    c = getc();
  }
  for (; c >= '0' && c <= '9'; c = getc()) {
    ret = (ret << 3) + (ret << 1) + (c ^ 48);
  }
  return pos ? ret : -ret;
}
const size_t N = 1 << 13;
const int MOD = 998244353;
int qpow(int base, int e) {
  int ret = 1;
  for (; e; e >>= 1) {
    if (e & 1) ret = (LL)ret * base % MOD;
    base = (LL)base * base % MOD;
  }
  return ret;
}
void fft(LL a[], int lg) {
  int n = 1 << lg;
  static int rev[N], rev_lg = -1;
  if (rev_lg != lg) {
    for (int i = 0; i < n; ++i) {
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg >> 1);
    }
    rev_lg = lg;
  }
  for (int i = 0; i < n; ++i) {
    if (rev[i] > i) {
      swap(a[i], a[rev[i]]);
    }
  }
  for (int m = 1, l = 2; m < n; m <<= 1, l <<= 1) {
    static int w[N];
    w[0] = 1;
    int ww = qpow(3, (MOD - 1) / l);
    for (int i = 1; i < m; ++i) {
      w[i] = (LL)w[i - 1] * ww % MOD;
    }
    LL x, y, *xx = a, *yy = a + m;
    for (int i = 0, j; i < n; i += l) {
      for (j = 0; j < m; ++j, ++xx, ++yy) {
        x = *xx;
        y = *yy % MOD * w[j];
        *xx = x + y;
        *yy = x - y;
      }
      xx += m;
      yy += m;
    }
    if (l >> 15 & 1) {
      for (int i = 0; i < n; ++i) {
        a[i] %= MOD;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    a[i] %= MOD;
    if (a[i] < 0) {
      a[i] += MOD;
    }
  }
}
int fac[N], inv[N];
LL a[N], b[N];
int main() {
  int n = read();
  int m = n << 1 | 1;
  fac[0] = 1;
  for (int i = 1; i <= m; ++i) {
    fac[i] = (LL)fac[i - 1] * i % MOD;
  }
  inv[m] = qpow(fac[m], MOD - 2);
  for (int i = m; i >= 1; --i) {
    inv[i - 1] = (LL)inv[i] * i % MOD;
  }
  int lg = 0;
  for (; (1 << lg) < m; ++lg) {
  }
  for (int i = read(); i <= n; ++i) {
    a[i] = inv[i];
  }
  for (int i = 0; i <= n; ++i) {
    b[i] = i & 1 ? MOD - inv[i] : inv[i];
  }
  fft(a, lg);
  fft(b, lg);
  for (int i = 0, _ = 1 << lg; i < _; ++i) {
    (a[i] *= b[i]) %= MOD;
  }
  reverse(a + 1, a + (1 << lg));
  fft(a, lg);
  int ans = 0, pw = 1;
  for (int i = 0; i <= n; ++i) {
    ans = (ans + a[i] * fac[i] % MOD * fac[i] % MOD * inv[i << 1 | 1] % MOD *
                     pw % MOD * inv[n - i]) %
          MOD;
    if ((pw <<= 1) >= MOD) {
      pw -= MOD;
    }
  }
  cout << ans * read() % MOD * fac[n] % MOD * qpow(2, MOD - 1 - lg) % MOD;
  return 0;
}