#include <bits/stdc++.h>
using namespace std;
const long long mxsz = 4e6 + 5;
const long long mod = 998244353;
long long n, k;
long long d[15];
long long arr[mxsz], r[mxsz];
long long fpow(long long b, long long e) {
  long long ret = 1LL;
  for (; e > 0; e >>= 1) {
    if (e & 1) {
      ret *= b;
      ret %= mod;
    }
    b *= b;
    b %= mod;
  }
  return ret;
}
void ntt(long long *a, long long N, long long root) {
  for (long long i = 0; i < N; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
  for (long long i = 1; i < N; i <<= 1) {
    long long r = fpow(root, (mod - 1) / (i << 1));
    for (long long j = 0; j < N; j += (i << 1)) {
      for (long long k = 0, t = 1; k < i; k++, t = (t * r) % mod) {
        long long x1 = a[j + k], x2 = (a[i + j + k] * t) % mod;
        a[j + k] = (x1 + x2) % mod;
        a[i + j + k] = (x1 + mod - x2) % mod;
      }
    }
  }
}
int32_t main() {
  cin >> n >> k;
  long long mx = 0;
  for (long long i = 0; i < k; i++) {
    cin >> d[i];
    mx = max(mx, d[i]);
    arr[d[i]] = 1;
  }
  n /= 2;
  long long N = 1, sh = -1;
  while (N <= (n * 10)) {
    N <<= 1;
    sh++;
  }
  for (long long i = 0; i < N; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << sh);
  ntt(arr, N, 3);
  for (long long i = 0; i < N; i++) arr[i] = fpow(arr[i], n);
  ntt(arr, N, (mod + 1) / 3);
  long long inv = fpow(N, mod - 2);
  long long ans = 0;
  for (long long i = 0; i < N; i++) arr[i] = (arr[i] * inv) % mod;
  for (long long i = 0; i < N; i++) {
    ans += (arr[i] * arr[i]) % mod;
    ans %= mod;
  }
  cout << ans << '\n';
  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;
const long long mxsz = 4e6 + 5;
const long long mod = 998244353;
long long n, k;
long long d[15];
long long arr[mxsz], r[mxsz];
long long fpow(long long b, long long e) {
  long long ret = 1LL;
  for (; e > 0; e >>= 1) {
    if (e & 1) {
      ret *= b;
      ret %= mod;
    }
    b *= b;
    b %= mod;
  }
  return ret;
}
void ntt(long long *a, long long N, long long root) {
  for (long long i = 0; i < N; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
  for (long long i = 1; i < N; i <<= 1) {
    long long r = fpow(root, (mod - 1) / (i << 1));
    for (long long j = 0; j < N; j += (i << 1)) {
      for (long long k = 0, t = 1; k < i; k++, t = (t * r) % mod) {
        long long x1 = a[j + k], x2 = (a[i + j + k] * t) % mod;
        a[j + k] = (x1 + x2) % mod;
        a[i + j + k] = (x1 + mod - x2) % mod;
      }
    }
  }
}
int32_t main() {
  cin >> n >> k;
  long long mx = 0;
  for (long long i = 0; i < k; i++) {
    cin >> d[i];
    mx = max(mx, d[i]);
    arr[d[i]] = 1;
  }
  n /= 2;
  long long N = 1, sh = -1;
  while (N <= (n * 10)) {
    N <<= 1;
    sh++;
  }
  for (long long i = 0; i < N; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << sh);
  ntt(arr, N, 3);
  for (long long i = 0; i < N; i++) arr[i] = fpow(arr[i], n);
  ntt(arr, N, (mod + 1) / 3);
  long long inv = fpow(N, mod - 2);
  long long ans = 0;
  for (long long i = 0; i < N; i++) arr[i] = (arr[i] * inv) % mod;
  for (long long i = 0; i < N; i++) {
    ans += (arr[i] * arr[i]) % mod;
    ans %= mod;
  }
  cout << ans << '\n';
  return 0;
}

```