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>
const int N = 4010;
const int moder = 998244353;
int comb[N][N];
int poly[N];
int inv[N];
int coe[N], power[N];
int main() {
  power[0] = 1;
  for (int i = 1; i < N; ++i) {
    power[i] = 2 * power[i - 1] % moder;
  }
  inv[1] = 1;
  for (int i = 2; i < N; ++i) {
    inv[i] = moder - 1ll * inv[moder % i] * (moder / i) % moder;
  }
  for (int i = 0; i < N; ++i) {
    comb[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % moder;
    }
  }
  int n, k, l;
  scanf("%d%d%d", &n, &k, &l);
  int ans = 0;
  poly[0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = N - 1; j; --j) {
      poly[j] = poly[j - 1];
    }
    poly[0] = 0;
    for (int j = N - 1; j; --j) {
      poly[j] = (1ll * poly[j - 1] * (moder - 1) + poly[j]) % moder;
    }
    int sum = 1ll * power[i] * comb[n][i] % moder;
    for (int j = 0; j < N - 1; ++j) {
      coe[i] = (coe[i] + 1ll * inv[j + 1] * poly[j] % moder * sum) % moder;
    }
  }
  for (int i = k; i <= n; ++i) {
    int sum = 0;
    for (int j = i; j <= n; ++j) {
      sum = (sum + (j - i & 1 ? -1ll : 1ll) * comb[j][i] * coe[j]) % moder;
    }
    sum += sum < 0 ? moder : 0;
    ans = (ans + sum) % moder;
  }
  ans = 1ll * ans * l % moder;
  printf("%d\n", ans);
  return 0;
}