#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize("unroll-loops")
using namespace std;
int n, k;
int f[1000003], g[1000003], inv[1000003];
long long ans = 1;
int main() {
  int x, siz = 0, t = 0;
  scanf("%d%d", &n, &k);
  if (k == 1) {
    putchar('1');
    return 0;
  }
  n >>= 1;
  for (register int i = 1; i <= k; ++i) {
    scanf("%d", &x);
    f[x] = 1, t = max(t, x);
  }
  for (register int i = 0; i <= t && f[i] == 0; ++i) siz = max(siz, i + 1);
  if (siz > 0) {
    for (register int i = 0; i <= t; ++i) f[i] = f[i + siz];
    t -= siz;
  }
  inv[1] = g[0] = 1;
  int tt = t * n;
  for (register int i = 2; i <= tt; ++i)
    inv[i] =
        (long long)(998244353 - 998244353 / i) * inv[998244353 % i] % 998244353;
  for (register int i = 0; i != tt; ++i) {
    int tmp = 0;
    for (register int j = 0; j <= min(i, t - 1); ++j)
      tmp = (tmp + (long long)(j + 1) * f[j + 1] % 998244353 * g[i - j]) %
            998244353;
    tmp = (long long)tmp * n % 998244353;
    for (register int j = max(0, i - t); j != i; ++j)
      tmp = (tmp - (long long)(j + 1) * g[j + 1] * f[i - j]) % 998244353;
    g[i + 1] = (long long)tmp * inv[i + 1] % 998244353;
    ans += (long long)g[i + 1] * g[i + 1] % 998244353;
  }
  printf("%lld", ans % 998244353);
  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>
#pragma GCC optimize(2)
#pragma GCC optimize("unroll-loops")
using namespace std;
int n, k;
int f[1000003], g[1000003], inv[1000003];
long long ans = 1;
int main() {
  int x, siz = 0, t = 0;
  scanf("%d%d", &n, &k);
  if (k == 1) {
    putchar('1');
    return 0;
  }
  n >>= 1;
  for (register int i = 1; i <= k; ++i) {
    scanf("%d", &x);
    f[x] = 1, t = max(t, x);
  }
  for (register int i = 0; i <= t && f[i] == 0; ++i) siz = max(siz, i + 1);
  if (siz > 0) {
    for (register int i = 0; i <= t; ++i) f[i] = f[i + siz];
    t -= siz;
  }
  inv[1] = g[0] = 1;
  int tt = t * n;
  for (register int i = 2; i <= tt; ++i)
    inv[i] =
        (long long)(998244353 - 998244353 / i) * inv[998244353 % i] % 998244353;
  for (register int i = 0; i != tt; ++i) {
    int tmp = 0;
    for (register int j = 0; j <= min(i, t - 1); ++j)
      tmp = (tmp + (long long)(j + 1) * f[j + 1] % 998244353 * g[i - j]) %
            998244353;
    tmp = (long long)tmp * n % 998244353;
    for (register int j = max(0, i - t); j != i; ++j)
      tmp = (tmp - (long long)(j + 1) * g[j + 1] * f[i - j]) % 998244353;
    g[i + 1] = (long long)tmp * inv[i + 1] % 998244353;
    ans += (long long)g[i + 1] * g[i + 1] % 998244353;
  }
  printf("%lld", ans % 998244353);
  return 0;
}

```