#include <bits/stdc++.h>
constexpr int P = 998244353;
int power(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) res = 1ll * res * a % P;
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, k;
  std::cin >> n >> k;
  n /= 2;
  std::vector<int> a(k);
  for (int i = 0; i < k; ++i) std::cin >> a[i];
  std::sort(a.begin(), a.end());
  for (int i = k - 1; i >= 0; --i) a[i] -= a[0];
  std::vector<int> f(a.back() * n + 1);
  f[0] = 1;
  for (int i = 0; i < a.back() * n; ++i) {
    int res = 0;
    for (auto j : a)
      if (j && j <= i + 1) res = (res + 1ll * j * f[i - j + 1]) % P;
    res = 1ll * res * n % P;
    for (auto j : a)
      if (j && j <= i) res = (res + 1ll * (i - j + 1) * (P - f[i - j + 1])) % P;
    f[i + 1] = 1ll * res * power(i + 1, P - 2) % P;
  }
  int ans = 0;
  for (int i = 0; i <= a.back() * n; ++i) ans = (ans + 1ll * f[i] * f[i]) % P;
  std::cout << ans << "\n";
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

All bus tickets in Berland have their numbers. A number consists of n digits (n is even). Only k decimal digits d_1, d_2, ..., d_k can be used to form ticket numbers. If 0 is among these digits, then numbers may have leading zeroes. For example, if n = 4 and only digits 0 and 4 can be used, then 0000, 4004, 4440 are valid ticket numbers, and 0002, 00, 44443 are not.

A ticket is lucky if the sum of first n / 2 digits is equal to the sum of remaining n / 2 digits. 

Calculate the number of different lucky tickets in Berland. Since the answer may be big, print it modulo 998244353.

Input

The first line contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 10) — the number of digits in each ticket number, and the number of different decimal digits that may be used. n is even.

The second line contains a sequence of pairwise distinct integers d_1, d_2, ..., d_k (0 ≤ d_i ≤ 9) — the digits that may be used in ticket numbers. The digits are given in arbitrary order.

Output

Print the number of lucky ticket numbers, taken modulo 998244353.

Examples

Input

4 2
1 8


Output

6


Input

20 1
6


Output

1


Input

10 5
6 1 4 0 3


Output

569725


Input

1000 7
5 4 0 1 8 3 2


Output

460571165

Note

In the first example there are 6 lucky ticket numbers: 1111, 1818, 1881, 8118, 8181 and 8888.

There is only one ticket number in the second example, it consists of 20 digits 6. This ticket number is lucky, so the answer is 1.

### Response

```cpp
#include <bits/stdc++.h>
constexpr int P = 998244353;
int power(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) res = 1ll * res * a % P;
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, k;
  std::cin >> n >> k;
  n /= 2;
  std::vector<int> a(k);
  for (int i = 0; i < k; ++i) std::cin >> a[i];
  std::sort(a.begin(), a.end());
  for (int i = k - 1; i >= 0; --i) a[i] -= a[0];
  std::vector<int> f(a.back() * n + 1);
  f[0] = 1;
  for (int i = 0; i < a.back() * n; ++i) {
    int res = 0;
    for (auto j : a)
      if (j && j <= i + 1) res = (res + 1ll * j * f[i - j + 1]) % P;
    res = 1ll * res * n % P;
    for (auto j : a)
      if (j && j <= i) res = (res + 1ll * (i - j + 1) * (P - f[i - j + 1])) % P;
    f[i + 1] = 1ll * res * power(i + 1, P - 2) % P;
  }
  int ans = 0;
  for (int i = 0; i <= a.back() * n; ++i) ans = (ans + 1ll * f[i] * f[i]) % P;
  std::cout << ans << "\n";
  return 0;
}

```