#include <bits/stdc++.h>
using namespace std;
unsigned fr[1 << 21], fs;
unsigned long long inv(unsigned long long x) {
  return x == 1
             ? 1
             : ((inv(998244353 % x) * (998244353 - 998244353 / x) + 998244353) %
                998244353);
}
unsigned long long qmpow(unsigned long long w, unsigned long long x) {
  unsigned long long r = 1;
  for (unsigned long long i = 1; i && i <= x;
       i <<= 1, w = ((w * w + 998244353) % 998244353))
    if (i & x) r = ((r * w + 998244353) % 998244353);
  return r;
}
void init(unsigned s) {
  for (fs = 1; fs <= s; fs <<= 1)
    ;
  for (int i = 0; i != fs; ++i)
    fr[i] = (fr[i >> 1] >> 1) | (i & 1 ? (fs >> 1) : 0);
}
void ntt(unsigned long long* p, int f) {
  static const unsigned long long g = 3;
  for (int i = 0; i != fs; ++i)
    if (i < fr[i]) swap(p[i], p[fr[i]]);
  for (int i = 1; i != fs; i <<= 1) {
    unsigned long long e = (998244353 - 1) / (i << 1),
                       w0 = qmpow(g, f ? e : 998244353 - 1 - e);
    for (int j = 0; j != fs; j += (i << 1)) {
      unsigned long long w = 1;
      for (int k = 0; k != i; k++, w = ((w * w0 + 998244353) % 998244353)) {
        unsigned long long u = p[j + k],
                           v = ((w * p[i + j + k] + 998244353) % 998244353);
        p[j + k] = ((u + v + 998244353) % 998244353);
        p[i + j + k] = ((u - v + 998244353) % 998244353);
      }
    }
  }
}
void ntt_res(unsigned long long* p) {
  for (unsigned long long i = 0, is = inv(fs); i != fs; ++i)
    p[i] = ((p[i] * is + 998244353) % 998244353);
}
unsigned long long p0[1 << 21];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  unsigned n, k;
  cin >> n >> k;
  while (k--) {
    unsigned d;
    cin >> d;
    p0[d] = 1;
  }
  init(1 << 21 >> 1);
  ntt(p0, 1);
  for (unsigned i = 0; i != 1 << 21; ++i) p0[i] = qmpow(p0[i], n / 2);
  ntt(p0, 0);
  ntt_res(p0);
  unsigned long long res = 0;
  for (unsigned i = 0; i != 1 << 21; ++i)
    res = ((res + p0[i] * p0[i] + 998244353) % 998244353);
  cout << res << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
unsigned fr[1 << 21], fs;
unsigned long long inv(unsigned long long x) {
  return x == 1
             ? 1
             : ((inv(998244353 % x) * (998244353 - 998244353 / x) + 998244353) %
                998244353);
}
unsigned long long qmpow(unsigned long long w, unsigned long long x) {
  unsigned long long r = 1;
  for (unsigned long long i = 1; i && i <= x;
       i <<= 1, w = ((w * w + 998244353) % 998244353))
    if (i & x) r = ((r * w + 998244353) % 998244353);
  return r;
}
void init(unsigned s) {
  for (fs = 1; fs <= s; fs <<= 1)
    ;
  for (int i = 0; i != fs; ++i)
    fr[i] = (fr[i >> 1] >> 1) | (i & 1 ? (fs >> 1) : 0);
}
void ntt(unsigned long long* p, int f) {
  static const unsigned long long g = 3;
  for (int i = 0; i != fs; ++i)
    if (i < fr[i]) swap(p[i], p[fr[i]]);
  for (int i = 1; i != fs; i <<= 1) {
    unsigned long long e = (998244353 - 1) / (i << 1),
                       w0 = qmpow(g, f ? e : 998244353 - 1 - e);
    for (int j = 0; j != fs; j += (i << 1)) {
      unsigned long long w = 1;
      for (int k = 0; k != i; k++, w = ((w * w0 + 998244353) % 998244353)) {
        unsigned long long u = p[j + k],
                           v = ((w * p[i + j + k] + 998244353) % 998244353);
        p[j + k] = ((u + v + 998244353) % 998244353);
        p[i + j + k] = ((u - v + 998244353) % 998244353);
      }
    }
  }
}
void ntt_res(unsigned long long* p) {
  for (unsigned long long i = 0, is = inv(fs); i != fs; ++i)
    p[i] = ((p[i] * is + 998244353) % 998244353);
}
unsigned long long p0[1 << 21];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  unsigned n, k;
  cin >> n >> k;
  while (k--) {
    unsigned d;
    cin >> d;
    p0[d] = 1;
  }
  init(1 << 21 >> 1);
  ntt(p0, 1);
  for (unsigned i = 0; i != 1 << 21; ++i) p0[i] = qmpow(p0[i], n / 2);
  ntt(p0, 0);
  ntt_res(p0);
  unsigned long long res = 0;
  for (unsigned i = 0; i != 1 << 21; ++i)
    res = ((res + p0[i] * p0[i] + 998244353) % 998244353);
  cout << res << endl;
  return 0;
}

```