#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
long long powmod(long long x, long long y, long long m) {
  if (y == 0) return 1;
  if (y % 2) return (x * powmod(x, y - 1, m)) % m;
  return powmod((x * x) % m, y / 2, m);
}
long long inv(long long b, long long M) {
  long long u = 1, x = 0, s = b, t = M;
  while (s) {
    long long q = t / s;
    swap(x -= u * q, u);
    swap(t -= s * q, s);
  }
  return (x %= M) >= 0 ? x : x + M;
}
const int G = 3;
void fmt(vector<long long> &x, int sign = +1) {
  int n = x.size();
  for (int i = 0, j = 1; j < n - 1; ++j) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(x[i], x[j]);
  }
  long long h = powmod(G, (mod - 1) / n, mod);
  if (sign < 0) h = inv(h, mod);
  for (int m = 1; m < n; m *= 2) {
    long long w = 1, wk = powmod(h, n / (2 * m), mod);
    for (int i = 0; i < m; ++i) {
      for (int j = i; j < n; j += 2 * m) {
        long long u = x[j], d = (x[j + m] * w) % mod;
        x[j] = (u + d) % mod;
        x[j + m] = (u - d + mod) % mod;
      }
      w = w * wk % mod;
    }
  }
  if (sign < 0) {
    long long n_inv = inv(n, mod);
    for (auto &a : x) a = (a * n_inv) % mod;
  }
}
vector<long long> s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, n;
  int k;
  cin >> n >> k;
  int l = 2;
  while (l < 1000000) l <<= 1;
  s = vector<long long>(l, 0);
  for (int i = 1; i <= k; i++) {
    cin >> a;
    s[a] = 1;
  }
  fmt(s, 1);
  for (int i = 0; i < l; i++) s[i] = powmod(s[i], n >> 1, mod);
  fmt(s, -1);
  long long sol = 0;
  for (int i = 0; i < l; i++) {
    long long s1 = s[i];
    sol += (s1 * s1) % mod;
    sol %= mod;
  }
  cout << sol << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long mod = 998244353;
long long powmod(long long x, long long y, long long m) {
  if (y == 0) return 1;
  if (y % 2) return (x * powmod(x, y - 1, m)) % m;
  return powmod((x * x) % m, y / 2, m);
}
long long inv(long long b, long long M) {
  long long u = 1, x = 0, s = b, t = M;
  while (s) {
    long long q = t / s;
    swap(x -= u * q, u);
    swap(t -= s * q, s);
  }
  return (x %= M) >= 0 ? x : x + M;
}
const int G = 3;
void fmt(vector<long long> &x, int sign = +1) {
  int n = x.size();
  for (int i = 0, j = 1; j < n - 1; ++j) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(x[i], x[j]);
  }
  long long h = powmod(G, (mod - 1) / n, mod);
  if (sign < 0) h = inv(h, mod);
  for (int m = 1; m < n; m *= 2) {
    long long w = 1, wk = powmod(h, n / (2 * m), mod);
    for (int i = 0; i < m; ++i) {
      for (int j = i; j < n; j += 2 * m) {
        long long u = x[j], d = (x[j + m] * w) % mod;
        x[j] = (u + d) % mod;
        x[j + m] = (u - d + mod) % mod;
      }
      w = w * wk % mod;
    }
  }
  if (sign < 0) {
    long long n_inv = inv(n, mod);
    for (auto &a : x) a = (a * n_inv) % mod;
  }
}
vector<long long> s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, n;
  int k;
  cin >> n >> k;
  int l = 2;
  while (l < 1000000) l <<= 1;
  s = vector<long long>(l, 0);
  for (int i = 1; i <= k; i++) {
    cin >> a;
    s[a] = 1;
  }
  fmt(s, 1);
  for (int i = 0; i < l; i++) s[i] = powmod(s[i], n >> 1, mod);
  fmt(s, -1);
  long long sol = 0;
  for (int i = 0; i < l; i++) {
    long long s1 = s[i];
    sol += (s1 * s1) % mod;
    sol %= mod;
  }
  cout << sol << endl;
  return 0;
}

```