#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
long long a[2100000], r[2100000];
long long po(long long base, long long x) {
  long long ret = 1;
  while (x) {
    if (x & 1) ret = ret * base % MOD;
    base = base * base % MOD;
    x >>= 1;
  }
  return ret;
}
void ntt(int n, int type) {
  for (int i = 1; i < n; i++) {
    r[i] = (r[i >> 1] >> 1) | ((i & 1) * (n >> 1));
    if (i < r[i]) swap(a[i], a[r[i]]);
  }
  for (int i = 1; i < n; i <<= 1) {
    long long wn = po(3, type * (MOD / (i * 2)) + MOD - 1);
    for (int j = 0; j < n; j += i * 2) {
      long long w = 1;
      for (int k = 0; k < i; k++) {
        int x = a[j + k], y = w * a[j + k + i] % MOD;
        a[j + k] = (x + y) % MOD;
        a[j + k + i] = (x - y + MOD) % MOD;
        w = w * wn % MOD;
      }
    }
  }
  if (type == 1) return;
  int invn = po(n, MOD - 2);
  for (int i = 0; i < n; i++) a[i] = a[i] * invn % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  n >>= 1;
  for (int i = 0; i < k; i++) {
    int x;
    cin >> x;
    a[x] = 1;
  }
  int d = 1;
  while (d <= n * 18) d <<= 1;
  ntt(d, 1);
  for (int i = 0; i < d; i++) a[i] = po(a[i], n);
  ntt(d, -1);
  long long ans = 0;
  for (int i = 0; i <= n * 9; i++) ans = (ans + a[i] * a[i]) % MOD;
  cout << ans << "\n";
}

### 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;
const int MOD = 998244353;
long long a[2100000], r[2100000];
long long po(long long base, long long x) {
  long long ret = 1;
  while (x) {
    if (x & 1) ret = ret * base % MOD;
    base = base * base % MOD;
    x >>= 1;
  }
  return ret;
}
void ntt(int n, int type) {
  for (int i = 1; i < n; i++) {
    r[i] = (r[i >> 1] >> 1) | ((i & 1) * (n >> 1));
    if (i < r[i]) swap(a[i], a[r[i]]);
  }
  for (int i = 1; i < n; i <<= 1) {
    long long wn = po(3, type * (MOD / (i * 2)) + MOD - 1);
    for (int j = 0; j < n; j += i * 2) {
      long long w = 1;
      for (int k = 0; k < i; k++) {
        int x = a[j + k], y = w * a[j + k + i] % MOD;
        a[j + k] = (x + y) % MOD;
        a[j + k + i] = (x - y + MOD) % MOD;
        w = w * wn % MOD;
      }
    }
  }
  if (type == 1) return;
  int invn = po(n, MOD - 2);
  for (int i = 0; i < n; i++) a[i] = a[i] * invn % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  n >>= 1;
  for (int i = 0; i < k; i++) {
    int x;
    cin >> x;
    a[x] = 1;
  }
  int d = 1;
  while (d <= n * 18) d <<= 1;
  ntt(d, 1);
  for (int i = 0; i < d; i++) a[i] = po(a[i], n);
  ntt(d, -1);
  long long ans = 0;
  for (int i = 0; i <= n * 9; i++) ans = (ans + a[i] * a[i]) % MOD;
  cout << ans << "\n";
}

```