#include <bits/stdc++.h>
using namespace std;
const int Maxn = 10000005, p = 998244353;
int n, k, x[15], mini = 0x3f3f3f3f;
long long ans, P[15], dP[15], A[Maxn];
int get_inv(long long x) {
  if (x <= 1) return 1;
  return (p - p / x) * get_inv(p % x) % p;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) scanf("%d", &x[i]), mini = min(mini, x[i]);
  for (int i = 1; i <= k; i++) x[i] -= mini;
  for (int i = 1; i <= k; i++) {
    P[x[i]] = 1;
    if (x[i]) dP[x[i] - 1] = x[i];
  }
  A[0] = P[0];
  for (int i = 0; i < 5 * n; i++) {
    for (int j = 0; j < 10 && i - j >= 0; j++)
      (A[i + 1] += (n / 2) * A[i - j] % p * dP[j] % p) %= p;
    for (int j = 1; j < 10 && i - j + 1 >= 0; j++)
      (A[i + 1] += p - P[j] * A[i - j + 1] % p * (i - j + 1) % p) %= p;
    (A[i + 1] *= get_inv(P[0] * (i + 1) % p)) %= p;
    (ans += A[i + 1] * A[i + 1]) %= p;
  }
  printf("%lld", (ans + 1) % p);
  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>
using namespace std;
const int Maxn = 10000005, p = 998244353;
int n, k, x[15], mini = 0x3f3f3f3f;
long long ans, P[15], dP[15], A[Maxn];
int get_inv(long long x) {
  if (x <= 1) return 1;
  return (p - p / x) * get_inv(p % x) % p;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) scanf("%d", &x[i]), mini = min(mini, x[i]);
  for (int i = 1; i <= k; i++) x[i] -= mini;
  for (int i = 1; i <= k; i++) {
    P[x[i]] = 1;
    if (x[i]) dP[x[i] - 1] = x[i];
  }
  A[0] = P[0];
  for (int i = 0; i < 5 * n; i++) {
    for (int j = 0; j < 10 && i - j >= 0; j++)
      (A[i + 1] += (n / 2) * A[i - j] % p * dP[j] % p) %= p;
    for (int j = 1; j < 10 && i - j + 1 >= 0; j++)
      (A[i + 1] += p - P[j] * A[i - j + 1] % p * (i - j + 1) % p) %= p;
    (A[i + 1] *= get_inv(P[0] * (i + 1) % p)) %= p;
    (ans += A[i + 1] * A[i + 1]) %= p;
  }
  printf("%lld", (ans + 1) % p);
  return 0;
}

```