Problem: Nezzar buys his favorite snack — n chocolate bars with lengths l_1,l_2,…,l_n. However, chocolate bars might be too long to store them properly! 

In order to solve this problem, Nezzar designs an interesting process to divide them into small pieces. Firstly, Nezzar puts all his chocolate bars into a black box. Then, he will perform the following operation repeatedly until the maximum length over all chocolate bars does not exceed k.

  * Nezzar picks a chocolate bar from the box with probability proportional to its length x. 
  * After step 1, Nezzar uniformly picks a real number r ∈ (0,x) and divides the chosen chocolate bar into two chocolate bars with lengths r and x-r. 
  * Lastly, he puts those two new chocolate bars into the black box. 



Nezzar now wonders, what is the expected number of operations he will perform to divide his chocolate bars into small pieces.

It can be shown that the answer can be represented as P/Q, where P and Q are coprime integers and Q not ≡ 0 (mod 998 244 353). Print the value of P⋅ Q^{-1} mod 998 244 353.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50, 1 ≤ k ≤ 2000).

The second line contains n integers l_1, l_2, …, l_n (1 ≤ l_i, ∑_{i=1}^{n} l_i ≤ 2000).

Output

Print a single integer — the expected number of operations Nezzar will perform to divide his chocolate bars into small pieces modulo 998 244 353.

Examples

Input


1 1
2


Output


4


Input


1 1
1


Output


0


Input


1 5
1234


Output


15630811


Input


2 1
2 3


Output


476014684


Input


10 33
10 20 30 40 50 60 70 80 90 100


Output


675105648

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
#define P 998244353
#define maxN 2234

i64 dp[maxN][maxN], tmp[maxN][maxN], fac[maxN], rfac[maxN];
int n, k;
i64 fastpow(i64 a, int b) {
  i64 ans = 1;
  a %= P;
  while (b) {
    if (b % 2)
      ans = ans * a % P;
    a = a * a % P;
    b /= 2;
  }
  return ans;
}

int main() {
  scanf("%d%d", &n, &k);
  dp[0][0] = 1;
  fac[0] = rfac[0] = 1;
  for (int i = 1; i <= 2000; i++) {
    fac[i] = fac[i - 1] * i % P;
    rfac[i] = fastpow(fac[i], P - 2);
  }
  int s = 0;
  for (int i = 0; i < n; i++) {
    int l;
    scanf("%d", &l);
    s += l;
    vector<array<int, 3>> term;
    for (int j = 0; l - j * k > 0; j++) {
      int sign = 1;
      if (j % 2 == 1)
        sign = -1;
      term.push_back(
          {l - j * k, j, (int)(fastpow(l - j * k, j) * rfac[j] % P * sign)});
      if (j > 0)
        term.push_back(
            {l - j * k, j - 1,
             (int)(fastpow(l - j * k, j - 1) * rfac[j - 1] % P * sign)});
    }
    for (int p = 0; p <= 2000; p++)
      for (int q = 0; q <= 2000; q++)
        tmp[p][q] = dp[p][q], dp[p][q] = 0;
    for (int p = 0; p <= 2000; p++)
      for (int q = 0; q <= 2000; q++)
        if (tmp[p][q] != 0) {
          for (auto t : term)
            dp[p + t[0]][q + t[1]] =
                (dp[p + t[0]][q + t[1]] + tmp[p][q] * t[2]) % P;
        }
  }
  for (int i = 0; i <= 2000; i++) {
    for (int j = 0; j <= 2000; j++)
      dp[i][j] = (P - dp[i][j]) % P;
  }
  dp[s][0] = 0;
  i64 invs = fastpow(s, P - 2), ans = 0;
  for (int a = 0; a <= 2000; a++)
    for (int b = 0; b <= 2000; b++)
      if (dp[a][b] != 0) {
        i64 tmp = dp[a][b] * fac[b] % P * fastpow(invs, b) % P *
                  fastpow(P + 1 - a * invs % P, P - 2 - b) % P;
        ans = (ans + tmp) % P;
      }
  printf("%lld\n", ans);
}