#include <bits/stdc++.h>
using namespace std;
long long n, m, k, x, ans = 0, id[1048580], a[1048580], b[1048580];
inline long long pw(long long x, long long n) {
  long long y = 1;
  for (; n; n >>= 1, x = x * x % 998244353) y = (n & 1) ? y * x % 998244353 : y;
  return y;
}
inline void NTT(long long *x, int t) {
  for (int i = (1); i <= (n - 1); ++i)
    if (i < id[i]) swap(x[i], x[id[i]]);
  for (int h = 2; h <= n; h <<= 1) {
    long long wn = pw(
        3, t == 1 ? (998244353 - 1) / h : 998244353 - 1 - (998244353 - 1) / h);
    for (int j = 0; j < n; j += h) {
      long long w = 1, tmp;
      for (int k = (j); k <= (j + (h >> 1) - 1); ++k) {
        tmp = x[k + (h >> 1)] * w % 998244353;
        x[k + (h >> 1)] = (x[k] - tmp + 998244353) % 998244353;
        x[k] = (x[k] + tmp) % 998244353;
        w = w * wn % 998244353;
      }
    }
  }
  if (t == -1) {
    long long invn = pw(n, 998244353 - 2);
    for (int i = (0); i <= (n - 1); ++i) x[i] = x[i] * invn % 998244353;
  }
}
int main() {
  cin >> m >> k;
  m /= 2;
  for (int i = (1); i <= (k); ++i) cin >> x, a[x] = 1;
  for (n = 1; n <= m * 9; n <<= 1)
    ;
  for (int i = (1); i <= (n - 1); ++i)
    id[i] = id[i >> 1] >> 1 | (i & 1 ? n >> 1 : 0);
  NTT(a, 1);
  for (int i = (0); i <= (n - 1); ++i) a[i] = pw(a[i], m);
  NTT(a, -1);
  for (int i = (0); i <= (n); ++i) ans = (ans + a[i] * a[i]) % 998244353;
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long n, m, k, x, ans = 0, id[1048580], a[1048580], b[1048580];
inline long long pw(long long x, long long n) {
  long long y = 1;
  for (; n; n >>= 1, x = x * x % 998244353) y = (n & 1) ? y * x % 998244353 : y;
  return y;
}
inline void NTT(long long *x, int t) {
  for (int i = (1); i <= (n - 1); ++i)
    if (i < id[i]) swap(x[i], x[id[i]]);
  for (int h = 2; h <= n; h <<= 1) {
    long long wn = pw(
        3, t == 1 ? (998244353 - 1) / h : 998244353 - 1 - (998244353 - 1) / h);
    for (int j = 0; j < n; j += h) {
      long long w = 1, tmp;
      for (int k = (j); k <= (j + (h >> 1) - 1); ++k) {
        tmp = x[k + (h >> 1)] * w % 998244353;
        x[k + (h >> 1)] = (x[k] - tmp + 998244353) % 998244353;
        x[k] = (x[k] + tmp) % 998244353;
        w = w * wn % 998244353;
      }
    }
  }
  if (t == -1) {
    long long invn = pw(n, 998244353 - 2);
    for (int i = (0); i <= (n - 1); ++i) x[i] = x[i] * invn % 998244353;
  }
}
int main() {
  cin >> m >> k;
  m /= 2;
  for (int i = (1); i <= (k); ++i) cin >> x, a[x] = 1;
  for (n = 1; n <= m * 9; n <<= 1)
    ;
  for (int i = (1); i <= (n - 1); ++i)
    id[i] = id[i >> 1] >> 1 | (i & 1 ? n >> 1 : 0);
  NTT(a, 1);
  for (int i = (0); i <= (n - 1); ++i) a[i] = pw(a[i], m);
  NTT(a, -1);
  for (int i = (0); i <= (n); ++i) ans = (ans + a[i] * a[i]) % 998244353;
  cout << ans << endl;
  return 0;
}

```