#include <bits/stdc++.h>
using namespace std;
const long long p = 998244353;
long long po(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 0) {
    long long u = po(a, b / 2);
    return (u * u) % p;
  } else {
    long long u = po(a, b - 1);
    return (a * u) % p;
  }
}
long long inv(long long x) { return po(x, p - 2); }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> v;
  for (long long i = 0; i < 10; ++i) v.push_back(0);
  for (long long i = 0; i < k; ++i) {
    long long x;
    cin >> x;
    v[x] = 1;
  }
  long long t = (n / 2);
  vector<long long> ans;
  long long lst = 9;
  while (!v[lst]) lst--;
  for (long long i = 0; i < 10; ++i) ans.push_back(0);
  ans.push_back(v[lst]);
  for (long long i = (lst * t - 1); i >= 0; i--) {
    long long sum1 = 0;
    for (long long j = 1; j <= lst; ++j) {
      sum1 += ((ans[ans.size() - j] * v[lst - j]) % p) * (i + j);
      sum1 %= p;
      sum1 -= ((ans[ans.size() - j] * v[lst - j]) % p) * (lst - j) * t;
      sum1 %= p;
    }
    sum1 %= p;
    sum1 += p;
    sum1 %= p;
    ans.push_back((sum1 * inv(lst * t - i)) % p);
  }
  long long res = 0;
  for (auto h : ans) {
    res += (h * h);
    res %= p;
  }
  cout << res;
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const long long p = 998244353;
long long po(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 0) {
    long long u = po(a, b / 2);
    return (u * u) % p;
  } else {
    long long u = po(a, b - 1);
    return (a * u) % p;
  }
}
long long inv(long long x) { return po(x, p - 2); }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> v;
  for (long long i = 0; i < 10; ++i) v.push_back(0);
  for (long long i = 0; i < k; ++i) {
    long long x;
    cin >> x;
    v[x] = 1;
  }
  long long t = (n / 2);
  vector<long long> ans;
  long long lst = 9;
  while (!v[lst]) lst--;
  for (long long i = 0; i < 10; ++i) ans.push_back(0);
  ans.push_back(v[lst]);
  for (long long i = (lst * t - 1); i >= 0; i--) {
    long long sum1 = 0;
    for (long long j = 1; j <= lst; ++j) {
      sum1 += ((ans[ans.size() - j] * v[lst - j]) % p) * (i + j);
      sum1 %= p;
      sum1 -= ((ans[ans.size() - j] * v[lst - j]) % p) * (lst - j) * t;
      sum1 %= p;
    }
    sum1 %= p;
    sum1 += p;
    sum1 %= p;
    ans.push_back((sum1 * inv(lst * t - i)) % p);
  }
  long long res = 0;
  for (auto h : ans) {
    res += (h * h);
    res %= p;
  }
  cout << res;
  return 0;
}

```